EXERCISE 4.21. Find the roots of the following quadratic equations by

1.	EXERCISE 4.21. Find the roots of the following quadratic equations by factorisation:x2-3x - 10=0(ii) 2x2 + x-6=0(ii) 2x2 + 7x+5V2 = 0(iv) 2x2 - x + - = 08(v) 100 x2-20x+1=0
Answer» Solution: $\\ \circ \: {\boxed{\tt\green{ Question \ 1}}}$ x² – 3x – 10 =0 Taking LHS, $\colon\implies$ x² – 5x + 2x – 10 $\colon\implies$ x(x – 5) + 2(x – 5) $\colon\implies$ (x – 5)(x + 2) The roots of x² – 3x – 10 = 0 are the values of x for which (x – 5)(x + 2) = 0 . As a Result x – 5 = 0 or x + 2 = 0 $\colon\implies$ x = 5 or x = -2 $\\ \circ \: {\boxed{\tt\purple{ Question \ 2}}}$ 2x² + x – 6 = 0 Taking LHS, $\colon\implies$ 2x² + 4x – 3x – 6 $\colon\implies$ 2x(x + 2) – 3(x + 2) $\colon\implies$ (x + 2)(2x – 3) The roots 2x² + x – 6=0 are the values of x for which (x – 5)(x + 2) = 0 As a Result x + 2 = 0 or 2x – 3 = 0 $\colon\implies$ x = -2 or ${\tt{x = \dfrac{3}{2} }}$ $\\ \circ \: {\boxed{\tt\blue{ Question \ 3}}}$ √2x² + 7x + 5√2=0 Taking LHS, $\colon\implies$ √2x² + 5x + 2x + 5√2 $\colon\implies$ x(√2x + 5) + √2(√2x + 5) $\colon\implies$ (√2x + 5)(x + √2) The roots of √2x² + 7x + 5√2 = 0 are the values of x for which (x – 5)(x + 2) = 0 As a Result √2x + 5 = 0 or x + √2 = 0 $\colon\implies$ x = -5/√2 or x = -√2 $\\ \circ \: {\boxed{\tt\red{ Question \ 4}}}$ 2x² – x + 1/8 = 0 Taking LHS, $\colon\implies$ 1/8 (16x² – 8X + 1) $\colon\implies$ 1/8 (16x² – 4x -4x + 1) $\colon\implies$ 1/8 (4x(4x – 1) -1(4x – 1)) $\colon\implies$ 1/8(4x – 1)² The roots of 2x² – x + 1/8 = 0, are the values of x for which (4x – 1)²= 0 As a Result (4x – 1) = 0 or (4x – 1) = 0 $\colon\implies$ x = 1/4 or x = 1/4 $\\ \circ \: {\boxed{\tt\pink{ Question \ 5}}}$ 100x² – 20X + 1=0 Taking LHS, $\colon\implies$ 100x² – 10x – 10x + 1 $\colon\implies$ 10x(10x – 1) -1(10x – 1) $\colon\implies$ (10x – 1)² The roots 100x² – 20x + 1=0, are the values of x for which (10x – 1)² = 0 $\colon\implies$ (10x – 1) = 0 or (10x – 1) = 0 $\colon {\tt\bold{ x = \dfrac{1}{10} \: Or \: x = \dfrac{1}{10} }} \\$

EXERCISE 4.21. Find the roots of the following quadratic equations by factorisation:x2-3x - 10=0(ii) 2x2 + x-6=0(ii) 2x2 + 7x+5V2 = 0(iv) 2x2 - x + - = 08(v) 100 x2-20x+1=0

Answer»

Solution:

$\\ \circ \: {\boxed{\tt\green{ Question \ 1}}}$

x² – 3x – 10 =0

Taking LHS,

$\colon\implies$ x² – 5x + 2x – 10

$\colon\implies$ x(x – 5) + 2(x – 5)

$\colon\implies$ (x – 5)(x + 2)

The roots of x² – 3x – 10 = 0 are the values of x for which (x – 5)(x + 2) = 0 .

As a Result

x – 5 = 0 or x + 2 = 0

$\colon\implies$ x = 5 or x = -2

$\\ \circ \: {\boxed{\tt\purple{ Question \ 2}}}$

2x² + x – 6 = 0

Taking LHS,

$\colon\implies$ 2x² + 4x – 3x – 6

$\colon\implies$ 2x(x + 2) – 3(x + 2)

$\colon\implies$ (x + 2)(2x – 3)

The roots 2x² + x – 6=0 are the values of x for which (x – 5)(x + 2) = 0

As a Result

x + 2 = 0 or 2x – 3 = 0

$\colon\implies$ x = -2 or ${\tt{x = \dfrac{3}{2} }}$

$\\ \circ \: {\boxed{\tt\blue{ Question \ 3}}}$

√2x² + 7x + 5√2=0

Taking LHS,

$\colon\implies$ √2x² + 5x + 2x + 5√2

$\colon\implies$ x(√2x + 5) + √2(√2x + 5)

$\colon\implies$ (√2x + 5)(x + √2)

The roots of √2x² + 7x + 5√2 = 0 are the values of x for which (x – 5)(x + 2) = 0

As a Result

√2x + 5 = 0 or x + √2 = 0

$\colon\implies$ x = -5/√2 or x = -√2

$\\ \circ \: {\boxed{\tt\red{ Question \ 4}}}$

2x² – x + 1/8 = 0

Taking LHS,

$\colon\implies$ 1/8 (16x² – 8X + 1)

$\colon\implies$ 1/8 (16x² – 4x -4x + 1)

$\colon\implies$ 1/8 (4x(4x – 1) -1(4x – 1))

$\colon\implies$ 1/8(4x – 1)²

The roots of 2x² – x + 1/8 = 0, are the values of x for which (4x – 1)²= 0

As a Result

(4x – 1) = 0 or (4x – 1) = 0

$\colon\implies$ x = 1/4 or x = 1/4

$\\ \circ \: {\boxed{\tt\pink{ Question \ 5}}}$

100x² – 20X + 1=0

Taking LHS,

$\colon\implies$ 100x² – 10x – 10x + 1

$\colon\implies$ 10x(10x – 1) -1(10x – 1)

$\colon\implies$ (10x – 1)²

The roots 100x² – 20x + 1=0, are the values of x for which (10x – 1)² = 0

$\colon\implies$ (10x – 1) = 0 or (10x – 1) = 0

$\colon {\tt\bold{ x = \dfrac{1}{10} \: Or \: x = \dfrac{1}{10} }} \\$