NCERT Solutions for Exercise 4.3 Class 10 Maths Chapter 4 - Quadratic Equations

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

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

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

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

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

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

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

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

Q1 (iii) Find the roots of the following quadratic equations, if they exist, by the method of completing the square $4x^2+4\sqrt{3}+3=0$

Answer:

Given equation: $4x^2+4\sqrt{3}+3=0$

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

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

Adding and subtracting $(\frac{\sqrt{3}}{2}{)}^2$ in the equation, we get

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

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

$\Rightarrow (x+\frac{\sqrt{3}}{2})=0or(x+\frac{\sqrt{3}}{2})=0$

Hence there are the same roots and equal:

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

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

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

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

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

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

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

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 : $a{x}^2+bx+c=0$ , where a, b, and c are arbitrary constants.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(iii) $4x^2+4\sqrt{3}x+3=0$

The general form of a quadratic equation is : $a{x}^2+bx+c=0$ , 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 : $a{x}^2+bx+c=0$ , 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: $x-\frac{1}{x}=3,x\ne 0$

Answer:

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

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: $a{x}^2+bx+c=0$ , 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: $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30},x\ne -4,7$

Answer:

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

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:

$\Rightarrow x=1or2$

Q4 The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is $\frac{1}{3}$ . Find the present age.

Answer:

Let the present age of Rehman be ${}^{\prime }{x}^{\prime }$ 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.

Q5 In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

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.

Q6 The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

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 $(90+30)m=120m$

Q7 The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

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

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

Therefore, the numbers are $18and12$ or $18and-12$ .

Q8 A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Answer:

Let the speed of the train be $xkm/hr.$

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

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

According to the question,

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

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

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 $40km/hr$ .

Q9 Two water taps together can fill a tank in $9\frac{3}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer:

Let the time taken by the smaller pipe to fill the tank be $xhr.$

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.

Q10 An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Answer:

Let the average speed of the passenger train be $xkm/hr$ .

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,

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

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

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

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 $33km/hr$ and

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

Q11 Sum of the areas of two squares is 468 m ² . If the difference of their perimeters is 24 m, find the sides of the two squares.

Answer:

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

And the perimeters will be: $4xand4y$ respectively.

Areas $x^{2}and{y}^2$ respectively.

It is given that,

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

$4x-4y=24m$ .................................(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 $(12m+6m)=18m$ .