Question

The lengths of the legs of an isosceles right triangle are x and
`√(6x+16)`

a. Find the length of the hypotenuse, in terms of x, in simplest radical form
b. Find the length of all three sides of the triangle, in simplest radical form.

Answer 1

Check the picture below. Let's recall is an isosceles, so it has twin sides and is a right-triangle.

`x^2+(√(6x+16))^2\implies √(x^2+6x+16)\qquad \impliedby hypotenuse \\\\[-0.35em] ~\dotfill\\\\ x~~ = ~~√(6x+16)\implies \stackrel{\textit{squaring both sides}}{x^2=6x+16}\implies x^2-6x-16=0 \\\\\\ (x-8)(x+2)=0\implies x= \begin{cases} 8 ~~ \textit{\LARGE \checkmark}\\\\ -2 ~~ \bigotimes \end{cases} \\\\[-0.35em] ~\dotfill`

`\underset{\textit{length of each side}}{\stackrel{√((8)^2 + 6(8) + 16)}{{\Large \begin{array}{llll} 8√(2) \end{array}}}\hspace{5em}\stackrel{x}{\text{\Large 8}}\hspace{5em}\stackrel{√(6(8) - 16)}{\text{\Large 8}}}`

now, let's notice, the negative value is a legitimate value, however for this case is invalid since the lengths of the triangle can't be negative.