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# Ataur's Responses in WyzAnt Answers

#### need help solving pythagorean quadratic...

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

+ more- less
Asked by Eric from Brooklyn, NY
00

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.

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