need help solving pythagorean quadratic...

Question

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

Answer 1

so since you have (2x+4)^2 and (2x+6)^2. Your first job is to take this solve and uncoil these two expressions first.

(2x+4)^2 can be solved using the squared for formula. 

(a+b)^2 = (a^2 + 2*a*b + b^2) = (a+b) * (a+b) 

Therefore (2x+4)^2 = (2x)^2 + 2*2x*4 + (4)^2

                              = 4x^2 + 16x + 16

and Therefore (2x+6)^2 = (2x)^2 + 2*2x*6 + (6)^2

                                   = 4x^2 + 24x + 36

Now that those polynomials are taken cared we have to put all together in the solution:

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

Its better to put everything to one side and make it equal to zero.

so, x^2 + 4x^2 + 24x + 36 - (4x^2 +16x +16) = 4x^2 + 16x + 16 - ( 4x^2 + 16x + 16)

    x^2 + 4x^2 + 24x + 36 - 4x^2 -16x -16 = 0

group the x^2 with each other and add them and do the same for the x's and the numbers.

so, x^2 + 4X^2 - 4x^2 + 24x - 16x + 36 -16 = 0

then you have, x^2 + 8x + 20 = 0 

Now solve this equation using factorization or the quadratic equation.

reply to this comment if you need further help. 

 

Answer 2

(2x + 6)² - x² = (2x + 4)²
4x² + 24x + 36 - x² = 4x² + 16x + 16
24x + 36 - x² = 16x + 16
0 = x² - 24x - 36 + 16x + 16
x² - 8x - 20 = 0
x₁ = (8 - √(64 + 80))/2 = -2 (does not satisfy given problem: distance can not be negative)
x₂ = (8 + √(64 + 80))/2 = 10 , thus  x = 10