Question

factorization of x²-42-8=0​

Answer 1

answer : x=5√2 , -5√2

Step-by-step explanation:

Subtract 8 from -42

x²-50=0

x²-50=0

x²=50

x²-50=0x²=50x=±√50

x²-50=0x²=50x=±√50x=±5√2

ANSWER : x=5√2 , -5√2

hope you're satisfied buddy :)

Answer 2

Answer:

$x^{2} -42-8=0\\x^{2}-50=(x+\sqrt{50})(x-\sqrt{50})$

Step-by-step explanation:

Using quadratic formula:

x = $\frac{-b+\sqrt{b^{2} -4*a*c}}{2*a}$ , $\frac{-b-\sqrt{b^{2} -4*a*c}}{2*a}$

Where in a general polynomial,

$ax^{2} + bx + c = 0$

Thus, for the polynomial in question, a = 1, b = 0, c = (-42-8) = -50. Substituting into the two expressions for x yields:

x = $\frac{-(0)+\sqrt{(0)^{2} -4*1*(-50)}}{2*1}$ , $\frac{-(0)-\sqrt{(0)^{2} -4*1*(-50)}}{2*1}$

$x = (+\sqrt{50} , -\sqrt{50})\\\\ x^{2} -42-8 = x^{2} -50 = (x+\sqrt{50})(x-\sqrt{50})$