Question 1080879

to solve a right triangle, use Pythagorean theorem {{{c^2=a^2+b^2}}}
 
If {{{a=x+1}}},{{{ b=x}}} and {{{c=5}}}, then we have:

{{{5^2=(x+1)^2+x^2}}}

{{{25=x^2+2x+1+x^2}}}

{{{0=2x^2+2x+1-25}}}

{{{2x^2+2x-24=0}}}.......simplify, divide by {{{2}}}

{{{x^2+x-12=0}}}....factor

{{{x^2-3x+4x-12=0}}}

{{{(x^2-3x)+(4x-12)=0}}}

{{{x(x-3)+4(x-3)=0}}}

{{{(x - 3) (x + 4) = 0}}}

solutions:

{{{(x - 3)  = 0}}}=>{{{x=3}}}
{{{(x +4)  = 0}}}=>{{{x=-4}}}...........since side of the triangle cannot be negative, disregard {{{x=-4}}}

so,{{{x=3}}}

then 
{{{a=x+1}}}->{{{a=3+1}}}->{{{highlight(a=4)}}}

{{{ b=x}}} ->{{{highlight(b=3)}}}

and {{{highlight(c=5)}}}