The pythagorean theorem of this problem x^2+(2x+6)^2=(2x+4)^2

Question

I need to find out the pythagorean theorm and the quadratic equation to solve by factoring and using the zero factor, use the zero factor property to solve each binomial, creating a compound equation

Answer 1

x^2+(2x+6)^2 = (2x+4)^2

Use the formula (a+b)^2 = a^2+2ab+b^2

x^2+4x^2+36+24x = 4x^2+16x+16

x^2+8x+20 = 0

To compare it quadratic equation then a = 1, b = 8, c = 20

Roots are x = -b±√(b^2-4ac)/2a

x = [-8±√(64-80)]/2

x = [-8±√-16]/2

x = (-8+4i)/2 and x = (-8-4i)/2

Zero factors are (-8+4i)/2,(-8-4i)/2.

Compound equations are substitute the x = (-8+4i)/2 and (-8+4i)/2

[(-8+4i)/2]^2+[2(-8+4i)/2+6]^2 = [2(-8+4i)/2+4]^2

[(-8-4i)/2]^2+[2(-8-4i)/2+6]^2 = [2(-8-4i)/2+4]^2