NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations

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# NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations

Edited By Saumya.Srivastava | Updated on Sep 05, 2023 07:24 PM IST

## NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations

Here students will get NCERT Solutions For Quadratic Equation class 10 which are created by expert team at Careers360 keeping in mind of latest syllabus of CBSE 2023-24. These solutions are very helpful for students to understand the basic concepts in a better way as these cover all the topics mentioned in syllabus comprehensive also include step by step explanation of each problem. When we equate the quadratic polynomial to zero, then we get a quadratic equation. NCERT Solutions for Class 10 Maths will help students to get indept understanding of the concepts and thus help students to strategise their preparation. They must complete the NCERT Class 10 Maths syllabus to the earliest so that they can revise in a strategic way. Each exercises of Quadratic Equation Class 10 are solved here in detailled manner.

Understanding quadratic equations chapter 4 maths class 10 is crucial as they appear in various real-life scenarios. Students must focus on mastering this chapter in the 2023-24 CBSE Syllabus to excel in Class 10 Math exams. NCERT solutions are valuable tools for comprehension and self-assessment. Regular practice with these solutions aids in addressing weaknesses. In mathematics, answers are either right or wrong, making concentration essential for achieving full marks.

## NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations - Important Formulae

f(x) = ax2 + bx + c, Where a, b, c ∈ R and a ≠ 0.

• (α, β) = [-b ± (b2 - 4ac)]/2a

• x = [-b ± D]/2a

• D = b2 - 4ac

Nature of Roots of Quadratic Equation:

• D > 0: Roots are real and distinct.

• D = 0: Roots are real and equal.

• D < 0: Roots are imaginary.

The sum of Roots:

• S = α + β = -b/a

Product of Roots:

• P = αβ = c/a

Quadratic Equation in terms of Roots:

• x2 - (α + β)x + αβ = 0

• One common root: (b1c2 - b2c1) / (c1a2 - c2a1) = (c1a2 - c2a1) / (a1b2 - a2b1)

• Both roots common: a1/a2 = b1/b2 = c1/c2

• ax2 + bx + c = 0 or [(x + b/2a)2 -D/(4a2)]

• If a > 0, minimum value = (4ac - b2)/4a at x = -b/(2a)

• If a < 0, maximum value = (4ac - b2)/4a at x = -b/(2a)

The sum of Roots of Cubic Equation:

• If α, β, γ are roots of the cubic equation: ax3 + bx2 + cx + d = 0

α + β + γ = -b/a.

αβ + βγ + γα = c/a.

αβγ = -d/a.

## NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations (Intext Questions and Exercise)

Quadratic Equation Class 10 Excercise: 4.1

We have L.H.S. $(x+1)^2 = x^2+2x+1$

Therefore, $(x+1)^2 = 2(x-3)$ can be written as:

$\Rightarrow x^2+2x+1 = 2x-6$

i.e., $x^2+7 = 0$

Or $x^2+0x+7 = 0$

This equation is of type: $ax^2+bx+c = 0$ .

Hence, the given equation is a quadratic equation.

Given equation $x^2 - 2x = (-2)(3-x)$ can be written as:

$\Rightarrow x^2 -2x = -6+2x$

i.e., $x^2-4x+6 = 0$

This equation is of type: $ax^2+bx+c = 0$ .

Hence, the given equation is a quadratic equation.

L.H.S. $(x-2)(x+1)$ can be written as:

$= x^2+x-2x-2 = x^2-x-2$

and R.H.S $(x-1)(x+3)$ can be written as:

$= x^2+3x-x-3 = x^2+2x-3$

$\Rightarrow x^2-x-2 = x^2+2x-3$

i.e., $3x-1 = 0$

The equation is of the type: $ax^2+bx+c = 0,a\neq0$ .

Hence, the given equation is not a quadratic equation since a=0.

L.H.S. $(x-3)(2x+1)$ can be written as:

$= 2x^2+x-6x-3 = 2x^2-5x-3$

and R.H.S $(x)(x+5)$ can be written as:

$= x^2+5x$

$\Rightarrow 2x^2-5x-3 = x^2+5x$

i.e., $x^2-10x-3 = 0$

This equation is of type: $ax^2+bx+c = 0,a\neq0$ .

Hence, the given equation is a quadratic equation.

L.H.S. $(2x-1)(x-3)$ can be written as:

$= 2x^2-6x-x+3 = 2x^2-7x+3$

and R.H.S $(x+5)(x-1)$ can be written as:

$=x^2-x+5x-5 = x^2+4x-5$

$\Rightarrow 2x^2-7x+3 = x^2+4x-5$

i.e., $x^2-11x+8 = 0$

This equation is of type: $ax^2+bx+c = 0,a \neq 0$ .

Hence, the given equation is a quadratic equation.

L.H.S. $x^2+3x+1$

and R.H.S $(x-2)^2$ can be written as:

$= x^2-4x+4$

$\Rightarrow x^2+3x+1 = x^2- 4x+4$

i.e., $7x-3 = 0$

This equation is NOT of type: $ax^2+bx+c = 0 , a\neq0$ .

Here a=0, hence, the given equation is not a quadratic equation.

L.H.S. $(x+2)^3$ can be written as:

$= x^3+8+6x(x+2) =x^3+6x^2+12x+8$

and R.H.S $2x(x^2-1)$ can be written as:

$= 2x^3-2x$

$\Rightarrow x^3+6x^2+12x+8 = 2x^3-2x$

i.e., $x^3-6x^2-14x-8 = 0$

This equation is NOT of type: $ax^2+bx+c = 0$ .

Hence, the given equation is not a quadratic equation.

L.H.S. $x^3 -4x^2 -x +1$ ,

and R.H.S $(x-2)^3$ can be written as:

$= x^3-6x^2+12x-8$

$\Rightarrow x^3-4x^2-x+1 = x^3-6x^2+12x-8$

i.e., $2x^2-13x+9=0$

This equation is of type: $ax^2+bx+c = 0$ .

Hence, the given equation is a quadratic equation.

Given the area of a rectangular plot is $528m^2$ .

Let the breadth of the plot be $'b'$ .

Then, the length of the plot will be: $= 2b +1$ .

Therefore the area will be:

$=b(2b+1)\ m^2$ which is equal to the given plot area $528m^2$ .

$\Rightarrow 2b^2+b = 528$

$\Rightarrow 2b^2+b - 528 = 0$

Hence, the length and breadth of the plot will satisfy the equation $2b^2+b - 528 = 0$

Given the product of two consecutive integers is $306.$

Let two consecutive integers be $'x'$ and $'x+1'$ .

Then, their product will be:

$x(x+1) = 306$

Or $x^2+x- 306 = 0$ .

Hence, the two consecutive integers will satisfy this quadratic equation $x^2+x- 306 = 0$ .

Let the age of Rohan be $'x'$ years.

Then his mother age will be: $'x+26'$ years.

After three years,

Rohan's age will be $'x+3'$ years and his mother age will be $'x+29'$ years.

Then according to question,

The product of their ages 3 years from now will be:

$\Rightarrow (x+3)(x+29) = 360$

$\Rightarrow x^2+3x+29x+87 = 360$ Or

$\Rightarrow x^2+32x-273 = 0$

Hence, the age of Rohan satisfies the quadratic equation $x^2+32x-273 = 0$ .

Let the speed of the train be $'s'$ km/h.

The distance to be covered by the train is $480\ km$ .

$\therefore$ The time taken will be

$=\frac{480}{s}\ hours$

If the speed had been $8\ km/h$ less, the time taken would be: $\frac{480}{s-8}\ hours$ .

Now, according to question

$\frac{480}{s-8} - \frac{480}{s} = 3$

$\Rightarrow \frac{480x - 480(x-8)}{(x-8)x} = 3$

$\Rightarrow 480x - 480x+3840 = 3(x-8)x$

$\Rightarrow 3840 = 3x^2-24x$

$\Rightarrow 3x^2 -24x-3840 = 0$

Dividing by 3 on both the side

$x^2 -8x-1280 = 0$

Hence, the speed of the train satisfies the quadratic equation $x^2 -8x-1280 = 0$

Quadratic Equation Class 10 Excercise: 4.2

Given the quadratic equation: $x^2 - 3x - 10 =0$

Factorization gives, $x^2 - 5x+2x - 10 =0$

$\Rightarrow x^2 - 5x+2x - 10 =0$

$\Rightarrow x(x-5) +2(x-5) =0$

$\Rightarrow (x-5)(x+2) =0$

$\Rightarrow x= 5\ or\ -2$

Hence, the roots of the given quadratic equation are $5\ and\ -2$ .

Given the quadratic equation: $2x^2 + x - 6 = 0$

Factorisation gives, $2x^2 +4x-3x - 6 = 0$

$\Rightarrow 2x(x+2) -3(x+2) =0$

$\Rightarrow (x+2)(2x-3) = 0$

Hence, the roots of the given quadratic equation are

$-2\ and\ \frac{3}{2}$

Given the quadratic equation: $\sqrt2x^2 + 7x + 5\sqrt2 = 0$

Factorization gives, $\sqrt2x^2 + 5x+2x + 5\sqrt2 = 0$

$\Rightarrow x(\sqrt2 x +5) +\sqrt2 (\sqrt 2 x +5)= 0$

$\Rightarrow (\sqrt2 x +5)(x+\sqrt{2}) = 0$

$\Rightarrow x=\frac{-5}{\sqrt 2 }\ or\ -\sqrt 2$

Hence, the roots of the given quadratic equation are

$\frac{-5}{\sqrt 2 }\ and\ -\sqrt 2$

Given the quadratic equation: $2x^2 -x + \frac{1}{8} = 0$

Solving the quadratic equations, we get

$16x^2-8x+1 = 0$

Factorization gives, $\Rightarrow 16x^2-4x-4x+1 = 0$

$\Rightarrow 4x(4x-1)-1(4x-1) = 0$

$\Rightarrow (4x-1)(4x-1) = 0$

$\Rightarrow x=\frac{1}{4}\ or\ \frac{1}{4}$

Hence, the roots of the given quadratic equation are

$\frac{1}{4}\ and\ \frac{1}{4}$

Given the quadratic equation: $100x^2 -20x +1 = 0$

Factorization gives, $100x^2 -10x-10x +1 = 0$

$\Rightarrow 10x(10x-1)-10(10x-1) = 0$

$\Rightarrow (10x-1)(10x-1) = 0$

$\Rightarrow x=\frac{1}{10}\ or\ \frac{1}{10}$

Hence, the roots of the given quadratic equation are

$\frac{1}{10}\ and\ \frac{1}{10}$ .

From Example 1 we get:

Equations:

(i) $x^2-45x+324 = 0$

Solving by factorization method:

Given the quadratic equation: $x^2-45x+324 = 0$

Factorization gives, $x^2-36x-9x+324 = 0$

$\Rightarrow x(x-36) - 9(x-36) = 0$

$\Rightarrow (x-9)(x-36) = 0$

$\Rightarrow x=9\ or\ 36$

Hence, the roots of the given quadratic equation are $x=9\ and \ 36$ .

Therefore, John and Jivanti have 36 and 9 marbles respectively in the beginning.

(ii) $x^2-55x+750 = 0$

Solving by factorization method:

Given the quadratic equation: $x^2-55x+750 = 0$

Factorization gives, $x^2-30x-25x+750 = 0$

$\Rightarrow x(x-30) -25(x-30) = 0$

$\Rightarrow (x-25)(x-30) = 0$

$\Rightarrow x=25\ or\ 30$

Hence, the roots of the given quadratic equation are $x=25\ and \ 30$ .

Therefore, the number of toys on that day was $30\ or\ 25.$

Let two numbers be x and y .

Then, their sum will be equal to 27 and the product equals 182.

$x+y = 27$ ...............................(1)

$xy =182$ .................................(2)

From equation (2) we have:

$y = \frac{182}{x}$

Then putting the value of y in equation (1), we get

$x+\frac{182}{x} = 27$

Solving this equation:

$\Rightarrow x^2-27x+182 = 0$

$\Rightarrow x^2-13x-14x+182 = 0$

$\Rightarrow x(x-13)-14(x-13) = 0$

$\Rightarrow (x-14)(x-13) = 0$

$\Rightarrow x = 13\ or\ 14$

Hence, the two required numbers are $13\ and \ 14$ .

Let the two consecutive integers be $'x'\ and\ 'x+1'.$

Then the sum of the squares is 365.

. $x^2+ (x+1)^2 = 365$

$\Rightarrow x^2+x^2+1+2x = 365$

$\Rightarrow x^2+x-182 = 0$

$\Rightarrow x^2 - 13x+14x+182 = 0$

$\Rightarrow x(x-13)+14(x-13) = 0$

$\Rightarrow (x-13)(x-14) = 0$

$\Rightarrow x =13\ or\ 14$

Hence, the two consecutive integers are $13\ and\ 14$ .

Let the length of the base of the triangle be $b\ cm$ .

Then, the altitude length will be: $b-7\ cm$ .

Given if hypotenuse is $13\ cm$ .

Applying the Pythagoras theorem; we get

$Hypotenuse^2 = Perpendicular^2 + Base^2$

So, $(13)^2 = (b-7)^2 +b^2$

$\Rightarrow 169 = 2b^2+49-14b$

$\Rightarrow 2b^2-14b-120 = 0$ Or $b^2-7b-60 = 0$

$\Rightarrow b^2-12b+5b-60 = 0$

$\Rightarrow b(b-12) + 5(b-12) = 0$

$\Rightarrow (b-12)(b+5) = 0$

$\Rightarrow b= 12\ or\ -5$

But, the length of the base cannot be negative.

Hence the base length will be $12\ cm$ .

Therefore, we have

Altitude length $= 12cm -7cm = 5cm$ and Base length $= 12\ cm$

Let the number of articles produced in a day $= x$

The cost of production of each article will be $=2x+3$

Given the total production on that day was $Rs.90$ .

Hence we have the equation;

$x(2x+3) = 90$

$2x^2+3x-90 = 0$

$\Rightarrow 2x^2+15x-12x-90 = 0$

$\Rightarrow x(2x+15) - 6(2x+15) = 0$

$\Rightarrow (2x+15)(x-6) = 0$

$\Rightarrow x =-\frac{15}{2}\ or\ 6$

But, x cannot be negative as it is the number of articles.

Therefore, $x=6$ and the cost of each article $= 2x+3 = 2(6)+3 = 15$

Hence, the number of articles is 6 and the cost of each article is Rs.15.

Class 10 Maths Chapter 4 Quadratic Equations Excercise: 4.3

Given equation: $2x^2 - 7x +3 = 0$

On dividing both sides of the equation by 2, we obtain

$\Rightarrow x^2-\frac{7}{2}x+\frac{3}{2} = 0$

$\Rightarrow (x-\frac{7}{4})^2 + \frac{3}{2} - \frac{49}{16} = 0$

$\Rightarrow (x-\frac{7}{4})^2 = \frac{49}{16} - \frac{3}{2}$

$\Rightarrow (x-\frac{7}{4})^2 =\frac{25}{16}$

$\Rightarrow (x-\frac{7}{4}) =\pm \frac{5}{4}$

$\Rightarrow x =\frac{7}{4}\pm \frac{5}{4}$

$\Rightarrow x = \frac{7}{4}+\frac{5}{4}\ or\ x = \frac{7}{4} - \frac{5}{4}$

$\Rightarrow x = 3\ or\ \frac{1}{2}$

Given equation: $2x^2 + x -4 = 0$

On dividing both sides of the equation by 2, we obtain

$\Rightarrow x^2+\frac{1}{2}x-2 = 0$

Adding and subtracting $\frac{1}{16}$ in the equation, we get

$\Rightarrow (x+\frac{1}{4})^2 -2 - \frac{1}{16} = 0$

$\Rightarrow (x+\frac{1}{4})^2 =2+\frac{1}{16}$

$\Rightarrow (x+\frac{1}{4})^2 = \frac{33}{16}$

$\Rightarrow (x+\frac{1}{4}) =\pm \frac{\sqrt{33}}{4}$

$\Rightarrow x =\pm \frac{\sqrt{33}}{4} -\frac{1}{4}$

$\Rightarrow x = \frac{\pm \sqrt{33} - 1}{4}$

$\Rightarrow x = \frac{ \sqrt{33} - 1}{4}\ or\ x = \frac{ -\sqrt{33} - 1}{4}$

Given equation: $4x^2 + 4\sqrt3 + 3 = 0$

On dividing both sides of the equation by 4, we obtain

$\Rightarrow x^2+\sqrt3x+\frac{3}{4} = 0$

Adding and subtracting $(\frac{\sqrt3}{2})^2$ in the equation, we get

$\Rightarrow (x+\frac{\sqrt3}{2})^2 +\frac{3}{4} - (\frac{\sqrt3}{2})^2 = 0$

$\Rightarrow (x+\frac{\sqrt3}{2})^2 = \frac{3}{4} - \frac{3}{4} = 0$

$\Rightarrow (x+\frac{\sqrt3}{2}) = 0\ or\ (x+\frac{\sqrt3}{2}) = 0$

Hence there are the same roots and equal:

$\Rightarrow x = \frac{-\sqrt3}{2}\ or\ \frac{-\sqrt3}{2}$

Given equation: $2x^2 + x + 4 = 0$

On dividing both sides of the equation by 2, we obtain

$\Rightarrow x^2+\frac{x}{2}+2 = 0$

Adding and subtracting $(\frac{1}{4})^2$ in the equation, we get

$\Rightarrow (x+\frac{1}{4})^2 +2- (\frac{1}{4})^2 = 0$

$\Rightarrow (x+\frac{1}{4})^2 = \frac{1}{16} -2 = \frac{-31}{16}$

$\Rightarrow (x+\frac{1}{4}) = \pm \frac{\sqrt{-31}}{4}$

$\Rightarrow x = \pm \frac{\sqrt{-31}}{4} - \frac{1}{4}$

$\Rightarrow x = \frac{\sqrt{-31}-1}{4} \ or\ x = \frac{-\sqrt{-31}-1}{4}$

Here the real roots do not exist (in the higher studies we will study how to find the root of such equations).

(i) $2x^2-7x+3 = 0$

The general form of a quadratic equation is : $ax^2+bx+c = 0$ , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

$a = 2,\ b = -7\ c = 3$

And the quadratic formula for finding the roots is:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values in the quadratic formula, we obtain

$\Rightarrow x= \frac{7 \pm \sqrt{49-24}}{4}$

$\Rightarrow x= \frac{7 \pm 5}{4}$

$\Rightarrow x= \frac{7 + 5}{4} = 3\ or\ x= \frac{7 - 5}{4} = \frac{1}{2}$

Therefore, the real roots are: $x =3,\ \frac{1}{2}$

(ii) $2x^2+x-4 = 0$

The general form of a quadratic equation is : $ax^2+bx+c = 0$ , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

$a = 2,\ b = 1\ c =-4$

And the quadratic formula for finding the roots is:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values in the quadratic formula, we obtain

$\Rightarrow x= \frac{-1 \pm \sqrt{1+32}}{4}$

$\Rightarrow x= \frac{-1 \pm \sqrt{33}}{4}$

$\Rightarrow x= \frac{-1 + \sqrt{33}}{4} \ or\ x= \frac{-1 - \sqrt{33}}{4}$

Therefore, the real roots are: $x = \frac{-1+\sqrt{33}}{4}\ or\ \frac{-1-\sqrt{33}}{4}$

(iii) $4x^2+4\sqrt3x+3 = 0$

The general form of a quadratic equation is : $ax^2+bx+c = 0$ , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

$a = 4,\ b = 4\sqrt{3}\ c =3$

And the quadratic formula for finding the roots is:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values in the quadratic formula, we obtain

$\Rightarrow x= \frac{-4\sqrt{3} \pm \sqrt{48-48}}{8}$

$\Rightarrow x= \frac{-4\sqrt{3} \pm 0}{8}$

Therefore, the real roots are: $x = \frac{-\sqrt{3}}{2}\ or\ \frac{-\sqrt{3}}{2}$

(iv) $2x^2+x+4 = 0$

The general form of a quadratic equation is : $ax^2+bx+c = 0$ , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

$a = 2,\ b = 1,\ c =4$

And the quadratic formula for finding the roots is:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values in the quadratic formula, we obtain

$\Rightarrow x= \frac{-1 \pm \sqrt{1-32}}{4}$

$\Rightarrow x= \frac{-1 \pm \sqrt{-31}}{4}$

Here the term inside the root is negative

Therefore there are no real roots for the given equation.

Given equation: $x - \frac{1}{x} = 3, x\neq 0$

So, simplifying it,

$\Rightarrow \frac{x^2-1}{x} = 3$

$\Rightarrow x^2-3x-1 = 0$

Comparing with the general form of the quadratic equation: $ax^2+bx+c = 0$ , we get

$a=1,\ b=-3,\ c=-1$

Now, applying the quadratic formula to find the roots:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow x= \frac{3 \pm \sqrt{9+4}}{2}$

$\Rightarrow x= \frac{3 \pm \sqrt{13}}{2}$

Therefore, the roots are

$\Rightarrow x = \frac{3+\sqrt{13}}{2}\ or\ \frac{3 - \sqrt{13}}{2}$

Given equation: $\frac{1}{x+4} - \frac{1}{x- 7} = \frac{11}{30},\ x\neq -4,7$

So, simplifying it,

$\Rightarrow \frac{x-7-x-4}{(x+4)(x-7)} = \frac{11}{30}$

$\Rightarrow \frac{-11}{(x+4)(x-7)} = \frac{11}{30}$

$\Rightarrow (x+4)(x-7) = -30$

$\Rightarrow x^2-3x-28 = -30$ or $\Rightarrow x^2-3x+2 = 0$

Can be written as:

$\Rightarrow x^2-x-2x+2 = 0$

$\Rightarrow x(x-1) -2(x-1) = 0$

$\Rightarrow (x-2)(x-1) = 0$

Hence the roots of the given equation are:

$\Rightarrow x = 1\ or\ 2$

Let the present age of Rehman be $'x'$ years.

Then, 3 years ago, his age was $(x-3)$ years.

and 5 years later, his age will be $(x+5)$ years.

Then according to the question we have,

$\frac{1}{(x-3)}+\frac{1}{(x+5)} = \frac{1}{3}$

Simplifying it to get the quadratic equation:

$\Rightarrow \frac{x+5+x-3}{(x-3)(x+5)} = \frac{1}{3}$

$\Rightarrow \frac{2x+2}{(x-3)(x+5)} = \frac{1}{3}$

$\Rightarrow 3(2x+2)= (x-3)(x+5)$

$\Rightarrow 6x+6 = x^2+2x-15$

$\Rightarrow x^2-4x-21 = 0$

$\Rightarrow x^2-7x+3x-21 = 0$

$\Rightarrow x(x-7)+3(x-7) = 0$

$\Rightarrow (x-7)(x+3) = 0$

Hence the roots are: $\Rightarrow x = 7,\ -3$

However, age cannot be negative

Therefore, Rehman is 7 years old in the present.

Let the marks obtained in Mathematics be 'm' then, the marks obtain in English will be '30-m'.

Then according to the question:

$(m+2)(30-m-3) = 210$

Simplifying to get the quadratic equation:

$\Rightarrow m^2-25m+156 = 0$

Solving by the factorizing method:

$\Rightarrow m^2-12m-13m+156 = 0$

$\Rightarrow m(m-12)-13(m-12) = 0$

$\Rightarrow (m-12)(m-13) = 0$

$\Rightarrow m = 12,\ 13$

We have two situations when,

The marks obtained in Mathematics is 12, then marks in English will be 30-12 = 18.

Or,

The marks obtained in Mathematics is 13, then marks in English will be 30-13 = 17.

Let the shorter side of the rectangle be x m.

Then, the larger side of the rectangle wil be $= (x+30)\ m$ .

Diagonal of the rectangle:

$= \sqrt{x^2+(x+30)^2}\ m$

It is given that the diagonal of the rectangle is 60m more than the shorter side.

Therefore,

$\sqrt{x^2+(x+30)^2} = x+60$

$\Rightarrow x^2+(x+30)^2 = (x+60)^2$

$\Rightarrow x^2+x^2+900+60x = x^2+3600+120x$

$\Rightarrow x^2-60x-2700 = 0$

Solving by the factorizing method:

$\Rightarrow x^2-90x+30x-2700 = 0$

$\Rightarrow x(x-90)+30(x-90)= 0$

$\Rightarrow (x+30)(x-90) = 0$

Hence, the roots are: $x = 90,\ -30$

But the side cannot be negative.

Hence the length of the shorter side will be: 90 m

and the length of the larger side will be $(90+30)\ m =120\ m$

Given the difference of squares of two numbers is 180.

Let the larger number be 'x' and the smaller number be 'y'.

Then, according to the question:

$x^2-y^2 = 180$ and $y^2 = 8x$

On solving these two equations:

$\Rightarrow x^2-8x =180$

$\Rightarrow x^2-8x -180 = 0$

Solving by the factorizing method:

$\Rightarrow x^2-18x+10x -180 = 0$

$\Rightarrow x(x-18)+10(x-18) = 0$

$\Rightarrow (x-18)(x+10) = 0$

$\Rightarrow x=18,\ -10$

As the negative value of x is not satisfied in the equation: $y^2 = 8x$

Hence, the larger number will be 18 and a smaller number can be found by,

$y^2 = 8x$ putting x = 18, we obtain

$y^2 = 144\ or\ y = \pm 12$ .

Therefore, the numbers are $18\ and\ 12$ or $18\ and\ -12$ .

Let the speed of the train be $x\ km/hr.$

Then, time taken to cover $360km$ will be:

$=\frac{360}{x}\ hr$

According to the question,

$\Rightarrow (x+5)\left ( \frac{360}{x}-1 \right ) = 360$

$\Rightarrow 360-x+\frac{1800}{x} - 5 = 360$

$\Rightarrow x^2+5x-1800 = 0$

Now, solving by the factorizing method:

$\Rightarrow x^2+45x-40x-1800 = 0$

$\Rightarrow x(x+45)-40(x+45) = 0$

$\Rightarrow (x-40)(x+45) = 0$

$\Rightarrow x = 40,\ -45$

However, the speed cannot be negative hence,

The speed of the train is $40\ km/hr$ .

Let the time taken by the smaller pipe to fill the tank be $x\ hr.$

Then, the time taken by the larger pipe will be: $(x-10)\ hr$ .

The fraction of the tank filled by a smaller pipe in 1 hour:

$= \frac{1}{x}$

The fraction of the tank filled by the larger pipe in 1 hour.

$= \frac{1}{x-10}$
Given that two water taps together can fill a tank in $9\frac{3}{8} = \frac{75}{8}$ hours.

Therefore,

$\Rightarrow \frac{1}{x}+\frac{1}{x-10} = \frac{8}{75}$

$\Rightarrow \frac{x-10+x}{x(x-10)} = \frac{8}{75}$

$\Rightarrow \frac{2x-10}{x(x-10)} = \frac{8}{75}$

$\Rightarrow 150x-750 = 8x^2-80x$

$\Rightarrow 8x^2-230x+750 = 0$

$\Rightarrow 8x^2-200x-30x+750 = 0$

$\Rightarrow 8x(x-25) - 30(x-25) = 0$

$\Rightarrow (x-25)(8x+30) = 0$

Hence the roots are $\Rightarrow x = 25,\ \frac{-30}{8}$

As time is taken cannot be negative:

Therefore, time is taken individually by the smaller pipe and the larger pipe will be $25$ and $25-10 =15$ hours respectively.

Let the average speed of the passenger train be $x\ km/hr$ .

Given the average speed of the express train $= (x+11)\ km/hr$

also given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.

Therefore,

$\Rightarrow \frac{132}{x} - \frac{132}{x+11} = 1$

$\Rightarrow 132\left [ \frac{x+11-x}{x(x+11)} \right ] = 1$

$\Rightarrow \frac{132\times11}{x(x+11)} = 1$

Can be written as quadratic form:

$\Rightarrow x^2+11x-1452 = 0$

$\Rightarrow x^2+44x-33x-1452 = 0$

$\Rightarrow x(x+44)-33(x+44)= 0$

$\Rightarrow (x+44)(x-33) = 0$

Roots are: $\Rightarrow x = -44,\ 33$

As the speed cannot be negative.

Therefore, the speed of the passenger train will be $33\ km/hr$ and

The speed of the express train will be $33+11 = 44\ km/hr$ .

Let the sides of the squares be $'x'\ and\ 'y'$ . (NOTE: length are in meters)

And the perimeters will be: $4x\ and\ 4y$ respectively.

Areas $x^2\ and\ y^2$ respectively.

It is given that,

$x^2 + y^2 = 468\ m^2$ .................................(1)

$4x-4y = 24\ m$ .................................(2)

Solving both equations:

$x-y = 6$ or $x= y+6$ putting in equation (1), we obtain

$(y+6)^2 +y^2 = 468$

$\Rightarrow 2y^2+36+12y = 468$

$\Rightarrow y^2+6y - 216 = 0$

Solving by the factorizing method:

$\Rightarrow y^2+18y -12y-216 = 0$

$\Rightarrow y(y+18) -12(y+18) = 0$

$\Rightarrow (y+18)(y-12)= 0$

Here the roots are: $\Rightarrow y = -18,\ 12$

As the sides of a square cannot be negative.

Therefore, the sides of the squares are $12m$ and $(12\ m+6\ m) = 18\ m$ .

Quadratic Equation Class 10 Excercise: 4.4

$2x^2 - 3x +5 = 0$

For a quadratic equation, $ax^2+bx+c = 0$ the value of discriminant determines the nature of roots and is equal to:

$D = b^2-4ac$

If D>0 then roots are distinct and real.

If D<0 then no real roots.

If D= 0 then there exists two equal real roots.

Given the quadratic equation, $2x^2 - 3x +5 = 0$ .

Comparing with general to get the values of a,b,c.

$a = 2, b =-3,\ c= 5$

Finding the discriminant:

$D= (-3)^2 - 4(2)(5) = 9-40 = -31$

$\because D<0$

Here D is negative hence there are no real roots possible for the given equation.

$b^2-4ac=(-4\sqrt{3})^2-(4\times4\times3)=48-48=0$

Here the value of discriminant =0, which implies that roots exist and the roots are equal.

The roots are given by the formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\sqrt{3}\pm\sqrt{0}}{2\times3}=\frac{2}{\sqrt{3}}$

So the roots are

$\frac{2}{\sqrt{3}},\ \frac{2}{\sqrt{3}}$

$2x^2 - 6x + 3 = 0$

The value of the discriminant

$b^2-4ac=(-6)^2-4\times2\times3=12$

The discriminant > 0. Therefore the given quadratic equation has two distinct real root

roots are

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-6\pm\sqrt{12}}{2\times2}=\frac{3}{2}\pm\frac{\sqrt{3}}{2}$

So the roots are

$\frac{3}{2}+\frac{\sqrt{3}}{2}, \frac{3}{2}-\frac{\sqrt{3}}{2}$

$2x^2 + kx + 3 = 0$

For two equal roots for the quadratic equation: $ax^2+bx+c =0$

The value of the discriminant $D= 0$ .

Given equation: $2x^2 + kx + 3 = 0$

Comparing and getting the values of a,b, and, c.

$a = 2, \ b = k,\ c = 3$

The value of $D = b^2-4ac = (k)^2 - 4(2)(3)$

$\Rightarrow (k)^2 = 24$

Or, $\Rightarrow k=\pm \sqrt{24} = \pm 2\sqrt{6}$

$kx(x-2) + 6 = 0$

For two equal roots for the quadratic equation: $ax^2+bx+c =0$

The value of the discriminant $D= 0$ .

Given equation: $kx(x-2) + 6 = 0$

Can be written as: $kx^2-2kx+6 = 0$

Comparing and getting the values of a,b, and, c.

$a = k, \ b = -2k,\ c = 6$

The value of $D = b^2-4ac = (-2k)^2 - 4(k)(6) = 0$

$\Rightarrow 4k^2 - 24k = 0$

$\Rightarrow 4k(k-6) = 0$

$\Rightarrow k= 0\ or\ 6$

But $k= 0$ is NOT possible because it will not satisfy the given equation.

Hence the only value of $k$ is 6 to get two equal roots.

Let the breadth of mango grove be $'b'$ .

Then the length of mango grove will be $'2b'$ .

And the area will be:

$Area = (2b)(b) = 2b^2$

Which will be equal to $800m^2$ according to question.

$\Rightarrow 2b^2 = 800m^2$

$\Rightarrow b^2 - 400 = 0$

Comparing to get the values of $a,b,c$ .

$a=1, \ b= 0 , \ c = -400$

Finding the discriminant value:

$D = b^2-4ac$

$\Rightarrow 0^2-4(1)(-400) = 1600$

Here, $D>0$

Therefore, the equation will have real roots.

And hence finding the dimensions:

$\Rightarrow b^2 - 400 = 0$

$\Rightarrow b = \pm 20$

As negative value is not possible, hence the value of breadth of mango grove will be 20m.

And the length of mango grove will be: $= 2\times10 = 40m$

Let the age of one friend be $x\ years.$

and the age of another friend will be: $(20-x)\ years.$

4 years ago, their ages were, $x-4\ years$ and $20-x-4 \ years$ .

According to the question, the product of their ages in years was 48.

$\therefore (x-4)(20-x-4) = 48$

$\Rightarrow 16x-64-x^2+4x= 48$

$\Rightarrow -x^2+20x-112 = 0$ or $\Rightarrow x^2-20x+112 = 0$

Now, comparing to get the values of $a,\ b,\ c$ .

$a = 1,\ b= -20,\ c =112$

Discriminant value $D = b^2-4ac = (-20)^2 -4(1)(112) = 400-448 = -48$

As $D<0$ .

Therefore, there are no real roots possible for this given equation and hence,

This situation is NOT possible.

Let us assume the length and breadth of the park be $'l'\ and\ 'b'$ respectively.

Then, the perimeter will be $P = 2(l+b) = 80$

$\Rightarrow l+b = 40\ or\ b = 40 - l$

The area of the park is:

$Area = l\times b = l(40-l) = 40l - l^2$

Given : $40l - l^2 = 400$

$l^2 - 40l +400 = 0$

Comparing to get the values of a, b and c.

The value of the discriminant $D = b^2-4ac$

$\Rightarrow = b^2-4ac = (-40)^2 - 4(1)(400) = 1600 -1600 = 0$

As $D = 0$ .

Therefore, this equation will have two equal roots.

And hence the roots will be:

$l =\frac{-b}{2a}$

$l =\frac{-40}{2(1)} = \frac{40}{2} =20$

Therefore, the length of the park, $l =20\ m$

and breadth of the park $b = 40-l = 40 -20 = 20\ m$ .

## Topics of Class 10 Maths Chapter 4 Quadratic Equations

• Representation of situation in a quadratic equation.

• Checking if an equation is quadratic or not.

• The solution of the quadratic equation using completing the square method.

• Solving a quadratic equation using the Sridharacharya formula.

• Product of roots in a quadratic equation.

• Sum of roots in a quadratic equation.

There are 4 exercise in Class 10 Maths chapter 4. Get the exercise wise solutions from the following links.

## Key Features For Chapter 4 Maths Class 10

Expertly Crafted: Skilled teachers at Careers360 create these NCERT Solutions with great care.

Accuracy Guaranteed: Class 10 maths chapter 4 solutions are entirely correct, making them perfect for students getting ready for their CBSE board exams.

Thorough Explanation: Even the smallest details are explained to help students feel more confident when facing other competitive exams.

Step-by-Step Answers: Class 10 maths ncert chapter 4 solutions to the textbook exercises are provided step by step. This helps students not only get the right final answers but also understand each part of the process, which can lead to better scores.

## NCERT solutions for class 10 Maths - Chapter Wise

 Chapter No. Chapter Name Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

### How To use Solutions Of Quadratic Equation Class 10 ?

• First of all list down all the questions in which you need assistance and go through the NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations of that particular question.

• When you complete the first step then your next target should be previous papers. You can pick past year papers and practice them thoroughly.

• Once you complete NCERTs and previous year papers, try to solve the questions of that particular chapter from different state board books.

### NCERT Exemplar solutions - Subject Wise

1. What are the important concepts learned in the NCERT Solutions for chapter 4 maths class 10 Quadratic Equations?

Class 10 maths chapter 4 solutions required multiple concepts including what is an equation, how to get the solution of an equation, methods of quadratic equation solution, the factoring method for solving a quadratic equation, the quadratic formula for solving a quadratic equation, the nature of roots of a quadratic equation, graphs of quadratic equations, and many.

2. What are the important topics covered in the NCERT Solutions for Class 10 Maths Chapter 4?

Chapter 4 quadratic equation class 10 ncert solutions contain multiple topics like Quadratic equations, Solutions of a quadratic equation, Discriminant of a quadratic equation, Factoring method for solving a quadratic equation, Quadratic formula for solving a quadratic equation, Nature of roots of a quadratic equation, and Graphs of quadratic equations.

Students can visit the NCERT official website. There they can find options to select class, subject, and book title. Using these inputs students can download NCERT chapter 4 maths class 10 which is quadratic equations class 10 pdf. Students can also find solutions to these NCERTs from Careers360's official website

4. How do you solve a quadratic equation in Class 10?

There are multiple methods to solve a quadratic equation including the factoring method, quadratic formula method, graphical method, and many others. These methods depend on the type of quadratic equation so first to know what is quadratic polynomial type and then apply the above methods to find a quadratic polynomial solution.

5. Which of the following is a quadratic equation?

To determine whether an equation is a quadratic equation, you can look for two conditions: firstly, the highest power of the variable is 2 and secondly, the coefficient of the x^2 term (that is, the "a" in the equation) is not equal to zero. These two are some examples of quadratic equations 2x^2+3x+5 and x^2+5x+6.

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##### Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
##### Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
##### Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story.

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion.

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
##### Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available
##### Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues.

5 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

2 Jobs Available
##### Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.

2 Jobs Available
##### Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
##### AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.

4 Jobs Available

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.

4 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process.

2 Jobs Available