Question 890770
{{{a^2}}}+{{{b^2}}}={{{c^2}}}
{{{16^2}}} + {{{(x+10)^2}}} = {{{(2x-6)^2}}}
256 + {{{x^2}}}+20x+100 = 4{{{x^2}}}-24x+36
Subtract {{{x^2}}} from each side
356 + 20x = 3{{{x^2}}}-24x+36
Subtract 356 from each side, subtract 20x from each side
0 = 3{{{x^2}}}-44x-320
Factor
(3x+16)(x-20)
X must be a counting number so we can disregard the first factor.
x-20=0
x=20
.
Let's plug it in to see if it works.
{{{a^2}}}+{{{b^2}}}={{{c^2}}}
Sides are 16, X=10 (20+10), hypotenuse 2x-6 (40-6)
{{{16^2}}}+{{{30^2}}}={{{34^2}}}
256+900=1156
Success!