Mathematics · Quadratic Equations

NCERT Examples

All 9 worked examples from Quadratic Equations, with independently verified solutions and direct navigation.

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Mathematics · Chapter 4 · NCERT Examples

Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.

NCERT Example

Example 1

forming quadratic equations · word problems

Represent the following situations mathematically.

John and Jivanti together have 45 marbles. After each loses 5 marbles, the product of the numbers left is 124.
A cottage industry makes x toys in a day. The cost per toy is ₹(55 − x), and the total cost is ₹750.

Solution

Let John initially have x marbles; Jivanti has 45 − x. After the losses:
$(x-5)(40-x)=124$
$-x^2+45x-324=0\Rightarrow x^2-45x+324=0$
Let x be the number of toys. Then:
$x(55-x)=750$
$x^2-55x+750=0$

Final answer

The required equations are x² − 45x + 324 = 0 and x² − 55x + 750 = 0.

NCERT Example

Example 2

quadratic equations · standard form

Check whether each equation is quadratic.

$(x-2)^2+1=2x-3$
$x(x+1)+8=(x+2)(x-2)$
$x(2x+3)=x^2+1$
$(x+2)^3=x^3-4$

Solution

$x^2-4x+5=2x-3\Rightarrow x^2-6x+8=0$
Quadratic.
$x^2+x+8=x^2-4\Rightarrow x+12=0$
Not quadratic.
$2x^2+3x=x^2+1\Rightarrow x^2+3x-1=0$
Quadratic.
$x^3+6x^2+12x+8=x^3-4\Rightarrow x^2+2x+2=0$
Quadratic after dividing by 6.

NCERT Example

Example 3

factorisation · roots

Find the roots of 2x² − 5x + 3 = 0 by factorisation.

Solution

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

=2x(x-1)-3(x-1)=(2x-3)(x-1)

2x-3=0 \text{or} x-1=0

x=\frac32 \text{or} x=1

Substitution in the original equation gives zero for both roots.

Final answer

x=1, \frac32

NCERT Example

Example 4

factorisation · roots

Find the roots of 6x² − x − 2 = 0.

Solution

6x^2-x-2=6x^2+3x-4x-2

=3x(2x+1)-2(2x+1)=(3x-2)(2x+1)

3x-2=0 \text{or} 2x+1=0

x=\frac23 \text{or} x=-\frac12

Each value satisfies the original quadratic.

Final answer

x=\frac23, -\frac12

NCERT Example

Example 5

factorisation · equal roots · radicals

\text{Find the roots of }3x^2-2\sqrt6x+2=0.

Solution

3x^2-2\sqrt6x+2=3x^2-\sqrt6x-\sqrt6x+2

=(\sqrt3x-\sqrt2)^2

\sqrt3x-\sqrt2=0\Rightarrow x=\sqrt{\frac23}

The repeated factor gives the same root twice; substitution gives 0.

Final answer

x=\sqrt{\frac23}, \sqrt{\frac23}

NCERT Example

Example 6

factorisation · word problem

Find the dimensions of a prayer hall whose breadth is x metres, length is 2x + 1 metres, and area is 300 m².

Solution

x(2x+1)=300\Rightarrow2x^2+x-300=0

2x^2-24x+25x-300=0

(x-12)(2x+25)=0

x=12 \text{or} x=-\frac{25}{2}

Breadth cannot be negative, so −12.5 is rejected. The length is 2(12) + 1 = 25 m.

Final answer

Breadth = 12 m and length = 25 m.

NCERT Example

Example 7

discriminant · nature of roots

Find the discriminant of 2x² − 4x + 3 = 0 and hence state the nature of its roots.

Solution

a=2,\quad b=-4,\quad c=3

D=b^2-4ac=(-4)^2-4(2)(3)=16-24=-8

Since D < 0, the equation has no real roots.

Final answer

D = −8; no real roots.

NCERT Example

Example 8

quadratic formula · discriminant · word problem

A pole is to be placed on the boundary of a circular park of diameter 13 m so that its distances from diametrically opposite gates differ by 7 m. Determine whether this is possible and find the distances.

A circular park with diameter AB, point P on the circumference, and right triangle APB. — Figure 4.2: the pole P and diametrically opposite gates A and B.

Solution

Let BP = x m, so AP = x + 7 m. Since angle APB is 90°:

(x+7)^2+x^2=13^2

x^2+7x-60=0

a=1,\quad b=7,\quad c=-60

D=7^2-4(1)(-60)=289

x=\frac{-7\pm\sqrt{289}}2=5 \text{or} -12

Distance cannot be negative, so −12 is rejected. Thus BP = 5 m and AP = 12 m; 5² + 12² = 13² verifies the geometry.

Final answer

Yes. The pole is 5 m from one gate and 12 m from the other.

NCERT Example

Example 9

discriminant · equal roots

\text{Find the discriminant and roots of }3x^2-2x+\frac13=0.

Solution

a=3,\quad b=-2,\quad c=\frac13

D=(-2)^2-4(3)\left(\frac13\right)=0

x=\frac{-b}{2a}=\frac{2}{6}=\frac13

D = 0, so this root is repeated; substitution in the original equation gives 0.

Final answer

x=\frac13, \frac13

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