Question

a) 3 (x^2-9)=0
b) 2 (x^2-72)=0
c) x^2+7x+12
d) x^2+21x+98​

Answer 1

3 (x² - 9) = 0

(x² - 9) = 0/3

(x² - 9) = 0

(x + 3)(x - 3) = 0

x - 3 =  0         x + 3 = 0

x = 3                x = -3

CS = (-3 : 3)

2 (x² - 72) = 0

(x² - 72) = 0/2

(x² - 72) = 0

x² = 72

$\sqrt{x^{2} } = +/- \sqrt{72} \\\\x = +/- \sqrt{72}$

CS =  $(-\sqrt{72} ; + \sqrt{72)}$

x² + 7x + 12

x            + 4

x            + 3

4x + 3 =  7x

(x+4)(x+3)

x² + 21x + 98

x             + 14

x             + 7

14x + 7x = 21x

(x +14)(x+7) ​

Eternamente axllxa