NCERT Solutions Class 10th Maths Chapter – 4 Quadratic Equations Examples

Example 1. Represent the following situations mathematically:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was 750. We would like to find out the number of toys produced on that day.

Solution: (i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45-x (Why?).
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45-x-5 /= 40-x
Therefore, their product = (x-5) (40-x)
= 40x – x² – 200 + 5x
= x²+45x-200
So, -x²+45x-200 = 124 (Given that product = 124)
i.e., -x²+45x-324 = 0
i.e., x²-45x + 324 0
Therefore, the number of marbles John had, satisfies the quadratic equation
x²-45x + 324 = 0
which is the required representation of the problem mathematically.

(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55-x
So, the total cost of production (in rupees) that day = x (55-x)
Therefore, x (55-x)= 750
i.e., 55x-x² = 750
i.e., -x²+55x-750 = 0
i.e., x²-55x+750 = 0
Therefore, the number of toys produced that day satisfies the quadratic equation
x²-55x+750= 0
which is the required representation of the problem mathematically.

Example 2. Check whether the following are quadratic equations:
(i) (x-2)²+1=2x-3
(ii) x(x+1)+8= (x+2) (x-2)
(iii) x (2x+3)= x²+1
(iv) (x+2)=x³-4

Solution:
(i) LHS = (x-2)² + 1 = x² – 4x + 4 + 1 = x²-4x+5
Therefore, (x-2)2+1=2x-3 can be rewritten as
x²-4x+5=2x-3
i.e., x²-6x+8=0
It is of the form ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(ii) Since x(x+1)+8 = x²+x+8 and (x+2)(x-2) = x²-4
Therefore, x²+x+8= x²-4
i.e., x+12= 0
It is not of the form ax² + bx + c = 0.
Therefore, the given equation is not a quadratic equation.

(iii) Here, LHS = x (2x+3)= 2x² + 3x
So, x (2x+3)= x² + 1 can be rewritten as
Therefore, we get x²+3x-1=0
2x² + 3x = x² + 1
It is of the form ax² + bx + c = 0.
So, the given equation is a quadratic equation.

(iv) Here, LHS = (x+2)³ = x²+6x² + 12x+8
Therefore, (x+2)³ = x² – 4 can be rewritten
x³+6x²+12x+8=x³-4
i.e., 6x² + 12x+12=0 or, x²+2x+2=0
It is of the form ax² + bx + c = 0.
So, the given equation is a quadratic equation.

Example 3. Find the roots of the equation 2x²-5x+3= 0, by factorisation.

Solution: Let us first split the middle term – 5x as -2x-3x [because (-2x) × (−3x) = 6x² = (2x²) × 3].
So, 2x²-5x+3=2x²-2x-3x+3= 2x (x − 1)-3(x-1) = (2x − 3)(x − 1)
Now, 2x²-5x+3= 0 can be rewritten as (2x-3)(x-1)=0.
So, the values of x for which 2x²-5x+3=0 are the same for which (2x-3)(x − 1) = 0,
i.e., either 2x-3=0 or x-1=0.
Now, 2x-3=0 gives x =3/2 and x-1=0 gives x = 1.
So, x=3/2 and x = 1 are the solutions of the equation.
In other words, 1 and are the roots of the equation 2x²-5x+3=0.

Example 4. Find the roots of the quadratic equation 6×2-x-2=0.

Solution: We have
6×2-x-2=6x²+3x-4x-2
= 3x (2x+1)-2(2x+1)
=(3x-2)(2x+1)
The roots of 6×2-x-2=0 are the values of x for which (3x-2)(2x+1)=0
Therefore, 3x-2= 0 or 2x + 1 = 0,
i.e., x =2/3 or x = -1/2
Therefore, the roots of 6×2-x-2= 0 are 2/3 and 1/2
We verify the roots, by checking that 2/3 and 1/2 satisfy 6×2-x-2=0.

Example 5. Find the roots of the quadratic equation 3x²-2√6x+2=0.

Solution: 3x²-2√6x+2 = 3x² -√6x-√6x+2
= √3x(√3x –√2) – √2 (√3x –√2)
= (√³ x − √2)(√3x −√2)
So, the roots of the equation are the values of x for which
(√3x-√2)(√3x-√2)=0
Now, √3x-√2=0 for x= √2/3
So, this root is repeated twice, one for each repeated factor √3x-√2.
Therefore, the roots of 3x² – 2√6x+2=0 are√2/3 √2/3