algebra equation, help??

x^4 - 15x^2 + 56 = 0

Look 56 = 7 x 8 & 15 = 7 + 8

so,
x^4 - (7 + 8)x^2 + 56 = 0

=> x^4 - 7x^2 - 8x^2 + 56 = 0

=> x^2(x^2 - 7) - 8 (x^2 - 7) = 0

=>(x^2 - 7)(x^2 - 8)= 0

=> x^2 - 7 = 0 or x^2 - 8 = 0

=> x = +/- sqrt(7) or x = +/- sqrt(8)

ANSWERS : x = +/- sqrt(7) , +/-sqrt(8)

(x^2-8)(x^2-7)=0
x^2-8=0, x^2-7=0
x^2=8, x^2=7
x=+-sqrt8, x=+-sqrt7
4 answers

make it a good day

x^4 - 15x^2 + 56 = 0
x^4 - 7x^2 - 8x^2 + 56 = 0
(x^4 - 7x^2) - (8x^2 - 56) = 0
x^2(x^2 - 7) - 8(x^2 - 7) = 0
(x^2 - 7)(x^2 - 8) = 0

x^2 - 7 = 0
x^2 = 7
x = ±√7

x^2 - 8 = 0
x^2 = 8
x = ±√8
x = 2(±√2)

∴ x = ±√7 , 2(±√2)

Take X^2 =Y solve for Y then X=+ or - 8^1/2 and + or - 7^1/2

This is a standard form of a quadratic equation:

Ax(squared) + Bx + C = 0

It is solved by using the quadratic formula; see source below.

x^4 - 15 x ² + 56 = 0
y ² - 15 y + 56 = 0
(y - 8)(y - 7) = 0
y = 8 , y = 7
x ² = 8 , x ² = 7
x = ± 2√2 , x = ±√7

(x^2-7)(x^2-8) = 0