Question

Quadratics Help please?

Answer 1

Remember that
for a right triangle

with legs a and b and hptonuse c
a²+b²=c²

given
a=x+5
b=x-3
c=9


(x+5)²+(x-3)²=9²
x²+10x+25+x²-6x+9=81
2x²+4x+34=81
2x²+4x-47=0

use quadratic formula

for 0=ax²+bx+c
x=$\frac{-b+/- \sqrt{b^2-4ac} }{2a}$
so
a=2
b=4
c=-47

x=$\frac{-4+/- \sqrt{4^2-4(2)(-47)} }{2(2)}$
x=$\frac{-4+/- \sqrt{16+376} }{4}$
x=$\frac{-4+/- \sqrt{392} }{4}$
x=$\frac{-4+/- 14\sqrt{2} }{4}$
x=$-1+/- 3.5\sqrt{2}$
x can't be negative because lengths can't be negative
so
x=-1+3.5√2
if we evaluate
x=3.94975
round
x=3.950 cm

Answer 2

(x-3)^2 + (x+5)^2=9^2
(x^2-6x+9) + (x^2+10x+25)=81
2x^2+4x+34=81
2x^2+4x-47=0
From here just use quadratic formula