Question

1.Find the roots of the following quadratic equation by factorization: a) 100x^2-20x+1=0 b)6x^2-x-2=0 c) 2x^2+x-300=0

Answer 1

Answer:

a.100x^2-20x-1

100x^2-10x-10x-1

10x(10x-1)-1(10x-1)

(10x-1)(10x-1)

b.6x^2-4x+3x-2

2x(3x-2)+1(3x-2)

(2x+1)(3x-2)

c.2x^2+25x-24x-300

X(2x+25)-12(2x+25)

(x-12)(2x+25)

Answer 2

Answer:

The roots are as follows :-

a) $\frac{1}{10} , \frac{1}{10}$

b) $\frac{2}{3} , \frac{-1}{2}$

c)$12 , \frac{25}{2}$

Step by step explanation:-

a) $100x^{2} - 20x + 1 = 0$

$100x^{2} - 10x - 10x +1 = 0\\\\10x (10x - 1) - 1(10x -1) = 0\\\\(10x - 1)(10x -1)= 0\\\\(10x - 1)^{2} = 0$

thus

$10x - 1 = \sqrt{0} \\\\10x - 1 = 0\\\\10x = 1$

$x = \frac{1}{10}$

b) $6x^{2} - x -2 = 0$

$6x^{2} - 4x + 3x - 2 = 0\\\\2x(3x - 2) + 1(3x - 2) = 0\\\\(3x - 2)(2x +1) = 0$

thus

$3x - 2 = 0\\\\3x = 2\\\\x = \frac{2}{3}$

& also

$2x + 1 =0\\\\2x = -1\\\\x = \frac{-1}{2}$

c) $2x^{2} + x -300 = 0$

$2x^{2} - 24x+ 25x - 300 = 0\\\\2x(x -12) + 25(x - 12) = 0\\\\(x -12)(2x - 25) = 0$

thus

$x - 12 = 0\\\\x = 12$

& also

$2x - 25 = 0\\\\2x = 25\\\\x = \frac{25}{2}$

I hope it helps........Please mark me as the brainliest