Question

Solve by factorisation method.

$1) \: \: 8 {x}^{2} - 22x - 21 = 0 \\ \\ 2)3 \sqrt{5} {x}^{2} + 25x + 10 \sqrt{5} = 0 \\ \\ 3)2 {x}^{2} + ax - {a}^{2} = 0 \\ \\ 4) \sqrt{3} {x}^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0$
#Give correct ans plz

Answer 1

(1):

8x² - 22x - 21 = 0

8x² - (28 - 6)x - 21 =0

8x² - 28x + 6x - 21 = 0

4x(2x - 7) + 3(2x - 7) = 0

(4x + 3)(2x - 7) = 0

By zero product rule,

(4x + 3 = 0) or (2x - 7 = 0)

(4x = -3) or (2x = 7)

$(x = - \frac{3}{4} ) \: \: \: \: or \: \: \: (x = \frac{7}{2} )$

(2):

3√5x² + 25x + 10√5 = 0

3√5x² + (15 + 10)x + 10√5 = 0

3√5x² + 15x + 10x + 10√5 = 0

3√5x(x + √5) + 10(x + √5) = 0

(x + √5)(3√5x + 10) = 0

By Zero Product Rule,

(x = -5) or (3√5x = -10)

$(x = - 5) \: \: \: \: \: or \: \: \: \: (x = - \frac{10}{3 \sqrt{5} } )$

By Rationalization,

$(x = - 5) \: \: \: \: \: or \: \: \: \: (x = - \frac{10 \sqrt{5} }{3 \sqrt{5} \times \sqrt{5} } = - \frac{10 \sqrt{5} }{15} = - \frac{2 \sqrt{5} }{3} )$

(3):

2x² + ax - a² = 0

2x² + (2 - 1)ax - a² = 0

2x² + 2ax - ax - a² = 0

2x(x + a) - a(x + a) = 0

(x + a) (2x - a) = 0

By Zero Product rule,

$x = - a \: \: \: \: \: or \: \: \: \: x = \frac{a}{2}$

(4):

√3x² - 2√2x -2√3 = 0

√3x² - (3√2 - 1√2)x - 2√3 = 0

√3x² - 3√2x + √2x - 2√3 = 0

√3x(x - √6) + √2(x - √6) = 0

(x - √6)(√3x + √2) = 0

By Zero Product rule,

( x = √6) or (x = -√2/√3)



There is a mistake in solution 2,

x = -√5 not -5
I hope this will help you

(-:

Answer 2

Hi sis!

Here is yr answer......

1) 8x² - 22x - 21 = 0

=> 8x² - 28x + 6x - 21 = 0

=> 4x(2x-7) + 3(2x-7) = 0

=> (2x-7)(4x+3) = 0

x = 7/2 or x = -3/4

2) 3√5x² + 25x + 10√5 = 0

=> 3√5x² + 15x + 10x + 10√5 = 0

=> 3√5x(x+√5) + 10(x+√5) = 0

=> (x+√5)(3√5x+10) = 0

x = -√5 or x = -10/3√5

3) 2x² + ax - a² = 0

=> 2x² + 2ax - ax - a² = 0

=> 2x(x+a) - a (x+a) = 0

=> (x+a)(2x-a) = 0

x = -a or x = a/2

4) √3x² - 2√2x - 2√3 = 0

=> √3x² - 3√2x + √2x - 2√3 = 0

=> √3x(x-√6) + √2(x-√6) = 0

=> (x-√6)(√3x+√2) = 0

x = √6 or x = -√2/√3

Hope it hlpz...