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Mathematics 11 Online
OpenStudy (anonymous):

Help, please? I'd love it if you showed me the steps. I want to use it for future reference. Solve: x^2 + 24^2 = (x + 8)^2

OpenStudy (whpalmer4):

\[x^2+24^2 = (x+8)^2\]I think the first step would be to expand the right hand side. Can you multiply \[(x+8)^2 = (x+8)(x+8) =\]and report back with the answer?

OpenStudy (anonymous):

x^2 + 576 = x^2 + 16x + 64 (expanding terms) Both sides have x^2 so they cancel 576 = 16x + 64 (cancelled out x^2's) 16x = 512 (subtract 576 from both sides) x = 32 (divide by 16 to isolate x)

OpenStudy (anonymous):

Oh, my gosh. I feel so dumb now. I knew you had to factor out (x + 8)^2 but my friend confused me saying you had to do quadratic formula in this problem and that's where I got stuck because x^2 cancelled each other out. Thank you so much for helping me. (:

OpenStudy (whpalmer4):

Yeah, no need for the quadratic formula here. A clever way you could do this would be to recognize that this has the same general form as the Pythagorean theorem: \[a^2+b^2=c^2\] 24 = 8*3 and 3 is the middle value of the Pythagorean triple \(3,4,5\) so we could immediately guess that a solution might be \(8*3, 8*4, 8*5\) which would allow us to have \(x=8*4=32\) and \(x+8 = 8*5= 40\) as the other two values... \[24^2+32^2 = 40^2\]\[576+1024 = 1600\] In case you have never thought of it in these terms, you can multiply a Pythagorean triple by any factor to scale it up: \[a^2+b^2= c^2\]\[(na)^2+(nb)^2=(nc)^2\]\[n^2a^2+n^2b^2=n^2c^2\]\[n^2(a^2+b^2) = n^2*c^2\]and of course that's equivalent to \[a^2+b^2 = c^2\]

OpenStudy (whpalmer4):

er, sorry, 3 is the first value of the Pythagorean triple, not the middle value. Anyhow, multiplying 3,4,5 by 8 spaces the values out to 24, 32, 40, and if x = 32, x+8 = 40, which is what we want.

OpenStudy (whpalmer4):

" had to factor out (x + 8)^2 " that's incorrect terminology, as you are not factoring, you are expanding (by multiplication). An example of factoring out would be this: \[x^3+2x^2+6x = x(x^2+2x+6)\]where you have "factored out" a factor of \(x\). Math is easier when you use the terminology the same way as everyone else :-)

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