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

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

Question:

Find the roots of the following quadratic equations by factorisation: √2x2 + 7x + 5√2 = 0

Given:

The quadratic equation: √2x2 + 7x + 5√2 = 0

To Find:

The values of x (roots) for the given quadratic equation.

Formula:

WKT, 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 quadratic equation is √2x2 + 7x + 5√2 = 0.
Comparing the given quadratic equation √2x2 + 7x + 5√2 = 0 with the standard form ax2 + bx + c = 0, we have:

a = √2, b = 7 and c = 5√2
Calculate the product (a × c) and sum (b):

Product (a × c) = √2 × 5√2 = 5 × √2 × √2 = 5 × 2 = 10

Sum (b) = 7
We need to find two numbers whose product is 10 and sum is 7.

Take the factors of the product without sign.

WKT, 10 = 2 x 5

The two numbers we want are m = 2 and n = 5.

Since product is 10 > 0 and sum is 7 > 0, both numbers are positive.

Therefore, m = 2 and n = 5
Split the middle term using m = 2 and n = 5:

⇒ √2x2 + 7x + 5√2 = 0

⇒ √2x2 + 2x + 5x + 5√2 = 0
Factor by grouping:

⇒ √2x2 + 2x + 5x + 5√2 = 0

⇒ √2x(x + √2) + 5(x + √2) = 0

⇒ (√2x + 5)(x + √2) = 0
Solve each factor:

From (√2x + 5)(x + √2) = 0

√2x + 5 = 0 or x + √2 = 0

Case 1: √2x + 5 = 0
⇒ √2x = -5
⇒ x = -5 √2

Rationalizing: x = -5√2 2

Case 2: x + √2 = 0
⇒ x = -√2

Result:

The roots of the quadratic equation √2x2 + 7x + 5√2 = 0 are x = -√2 and x = -5√2 2 .
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