Eric J.
asked 02/24/13need help solving pythagorean quadratic...
x^2+(2x+6)^2=(2x+4)^2
2 Answers By Expert Tutors
Ataur R. answered 02/24/13
Experienced in helping students with Math, Accounting and SAT tests.
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.
Eric J.
Thank you so much for your help Ataur, I see where I went wrong in my math. I now have to find two factors of 20 whose sum would equal the coefficient of 8, and there are none..
Can you help?
02/24/13
Nataliya D. answered 02/24/13
Patient and effective tutor for your most difficult subject.
(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
Eric J.
Thank you for your efforts Nataliya, they are greatly appreciated!
02/25/13
Nataliya D.
You"re very welcome Eric.
Sorry for late answer, I solved it before I wrote the note for you, but for some reason the answer did not appear. There was a note "Your answer is accepted", but it did not go through completely.
02/25/13
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Eric J.
Am I going to get an answer?
02/24/13