CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26
NCERT Solutions for Exercise 4.3 Class 10 Maths Chapter 4 Quadratic Equations are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This ex 4.3 class 10 deals concept of finding the roots of quadratic equations by completing squares and applying the quadratic formula. Exercise 4.3 Class 10 Maths, quadratic equations help us to explore the ideas of ways of finding quadratic equations in depth. Completing the square is nothing but finding the value that makes a quadratic equation a square trinomial. The square trinomial can then be solved easily by using algebraic identities and formulas. NCERT solutions for Class 10 Maths chapter 4 exercise 4.3 consists of 11 simple problems with word problems that are easy to solve and also explore the concepts of solving quadratic equations by using quadratic formulas.
NCERT solutions for Class 10 Maths exercise 4.3, focused on the concepts of solving quadratic equations and understanding more about the relationship between the roots of the equation and the nature of the equation. These class 10 maths ex 4.2 solutions are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.
Quadratic Equations class 10 chapter 4 Exercise: 4.3
Q1 (i) Find the roots of the following quadratic equations, if they exist, by the method of completing the square $2x^2 - 7x +3 = 0$
Answer:
Given equation: $2x^2 - 7x +3 = 0$
On dividing both sides of the equation by 2, we obtain
$\Rightarrow x^2-\frac{7}{2}x+\frac{3}{2} = 0$
$\Rightarrow (x-\frac{7}{4})^2 + \frac{3}{2} - \frac{49}{16} = 0$
$\Rightarrow (x-\frac{7}{4})^2 = \frac{49}{16} - \frac{3}{2}$
$\Rightarrow (x-\frac{7}{4})^2 =\frac{25}{16}$
$\Rightarrow (x-\frac{7}{4}) =\pm \frac{5}{4}$
$\Rightarrow x =\frac{7}{4}\pm \frac{5}{4}$
$\Rightarrow x = \frac{7}{4}+\frac{5}{4}\ or\ x = \frac{7}{4} - \frac{5}{4}$
$\Rightarrow x = 3\ or\ \frac{1}{2}$
Q1 (ii) Find the roots of the following quadratic equations, if they exist, by the method of completing the square $2x^2 + x -4 = 0$
Answer:
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
$\Rightarrow (x+\frac{1}{4})^2 -2 - \frac{1}{16} = 0$
$\Rightarrow (x+\frac{1}{4})^2 =2+\frac{1}{16}$
$\Rightarrow (x+\frac{1}{4})^2 = \frac{33}{16}$
$\Rightarrow (x+\frac{1}{4}) =\pm \frac{\sqrt{33}}{4}$
$\Rightarrow x =\pm \frac{\sqrt{33}}{4} -\frac{1}{4}$
$\Rightarrow x = \frac{\pm \sqrt{33} - 1}{4}$
$\Rightarrow x = \frac{ \sqrt{33} - 1}{4}\ or\ x = \frac{ -\sqrt{33} - 1}{4}$
Q1 (iii) Find the roots of the following quadratic equations, if they exist, by the method of completing the square $4x^2 + 4\sqrt3 + 3 = 0$
Answer:
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:
$\Rightarrow x = \frac{-\sqrt3}{2}\ or\ \frac{-\sqrt3}{2}$
Q2 (iv) Find the roots of the following quadratic equations, if they exist, by the method of completing the square $2x^2 + x + 4 = 0$
Answer:
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).
Q2 Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
Answer:
(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.
Q3 (i) Find the roots of the following equations: $x - \frac{1}{x} = 3, x\neq 0$
Answer:
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
$\Rightarrow x = \frac{3+\sqrt{13}}{2}\ or\ \frac{3 - \sqrt{13}}{2}$
Q3 (ii) Find the roots of the following equations: $\frac{1}{x+4} - \frac{1}{x- 7} = \frac{11}{30},\ x\neq -4,7$
Answer:
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:
$\Rightarrow x = 1\ or\ 2$
Answer:
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.
Answer:
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.
Answer:
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 $(90+30)\ m =120\ m$
Answer:
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 $18\ and\ -12$ .
Answer:
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,
$\Rightarrow (x+5)\left ( \frac{360}{x}-1 \right ) = 360$
$\Rightarrow 360-x+\frac{1800}{x} - 5 = 360$
Making it a quadratic equation.
$\Rightarrow x^2+5x-1800 = 0$
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 $40\ km/hr$ .
Answer:
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,
$\Rightarrow \frac{1}{x}+\frac{1}{x-10} = \frac{8}{75}$
$\Rightarrow \frac{x-10+x}{x(x-10)} = \frac{8}{75}$
$\Rightarrow \frac{2x-10}{x(x-10)} = \frac{8}{75}$
Making it a quadratic equation:
$\Rightarrow 150x-750 = 8x^2-80x$
$\Rightarrow 8x^2-230x+750 = 0$
$\Rightarrow 8x^2-200x-30x+750 = 0$
$\Rightarrow 8x(x-25) - 30(x-25) = 0$
$\Rightarrow (x-25)(8x+30) = 0$
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.
Answer:
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 $33+11 = 44\ km/hr$ .
Answer:
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 $(12\ m+6\ m) = 18\ m$ .
Exercise 4.3 Class 10 Maths consists of a question based on finding the roots of the Quadratic Equation by Completing the Square. In solving the Quadratic Equation by Completing the Square, first, we need to write the quadratic equation in general form that is ax2+bx+c=0 . For that we need to divide both sides of the equation by the coefficient of x2 if it not equal to 1 . Then we need to shift the constant term to the right-hand side. Then we need to add a square of one half on both sides. We need to write the left-hand side as the square and simplify the right-hand side. Take the square root of both sides to find the value of x. The NCERT solutions for Class 10 Maths exercise 4.3 also focused on the concepts of solving quadratic equations by using quadratic formulas. Few questions related to solving quadratic equations by using quadratic formulas are given also in exercise 4.3 Class 10 Maths. Also students can get access of Quadratic Equations Class 10 Notes to revise all the concepts quickly.
Also see-
Frequently Asked Questions (FAQs)
Yes, the given statement is true. A quadratic equation can be solved using the method of completing squares.
One of the fastest methods to calculate the roots of a quadratic equation is using the quadratic formula.
Completing the square is nothing but finding the value that makes a quadratic equation a square trinomial
The questions are centred on the concept of finding the roots of quadratic equations by completing squares, and the NCERT solutions for Class 10 Maths chapter 4 exercise 4.3 consists of 11 problems with subsections
On Question asked by student community
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