NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 1 (v)

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 1 (v)

Question:

Find the roots of the following quadratic equations by factorisation : 100x2 – 20x + 1 = 0

Given:

The quadratic equation: 100x2 – 20x + 1 = 0

To Find:

The roots of the equation by factorisation.

Formula:

For a quadratic equation ax² + bx + c = 0, we can find two numbers m and n such that:

                        ax² + bx + c = ax² + mx + nx + c

where

  • m × n = a × c
  • m + n = b

Cases:

  • If (a × c > 0) and b < 0, both m and n are negative.
  • If (a × c > 0) and b > 0, both m and n are positive.
  • If (a × c < 0), the numbers m and n have opposite signs.

Solution:

The given equation is 100x2 – 20x + 1 = 0.

Here, a = 100, b = -20, c = 1.
Product (a × c) = 100 × 1 = 100.

Sum (b) = -20.
We need to find two numbers whose product is 100 and sum is -20.

WKT, 100 = 10 x 10

The two numbers we want are m = 10 and n = 10.

Since product is 100 and 100 > 0 And Since sum is -20 and -20 < 0 both numbers are negative.

Therefore the two numbers are -10 and -10.
Rewrite the middle term (-20x) using these numbers:

⇒ 100x2 – 10x – 10x + 1 = 0
Factor by grouping:

⇒ 10x(10x – 1) – 1(10x – 1) = 0
Factor out the common binomial (10x – 1):

⇒ (10x – 1)(10x – 1) = 0
This is a perfect square trinomial, (10x – 1)2 = 0.
Set the factor equal to zero to find the root:

⇒ 10x – 1 = 0

⇒ 10x = 1

⇒ x = 1 10

Result:

The roots of the equation 100x2 – 20x + 1 = 0 are x = 1 10 (repeated root).
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