# NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

## Types of questions asked from class 10 maths chapter 4 Quadratic Equations

• Representation of situation in a quadratic equation.
• Checking of 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.

## NCERT solutions for class 10 maths chapter 4 Quadratic Equations 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 $x^2 -8x-1280 = 0$

## Q1 (i) Find the roots of the following quadratic equations by factorization: $x^2 - 3x - 10 =0$

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, $2x^2 +4x-3x - 6 = 0$

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

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

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

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.

## NCERT solutions for 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 $(12\ m+6\ m) = 18\ m$.

## NCERT solutions for class 10 maths chapter 4 Quadratic Equations 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 $b = 40-l = 40 -20 = 20\ m$.

## NCERT solutions for class 10 maths chapter wise

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

## How to use NCERT solutions for class 10 maths chapter 4 Quadratic Equations?

• First of all list down all the questions in which you need assistance and go through the NCERT solution 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.

Keep working hard & happy learning!