Answer:
[tex]x = \frac{-2+7\sqrt{2} }{2}[/tex] and [tex]\frac{-2-7\sqrt{2} }{2}[/tex]
Step-by-step explanation:
hey there,
< The way you solve right angle triangle problems are using the Pythagorean theorem. (I hope you have already learned it but if you haven't feel free to ask me.)
Pythagorean theorem states: [tex]a^2 + b^2 = c^2[/tex]
The two sides of the triangle that touch the right angle are (x+5) and (x-3).
Those two values count as "a" and "b". "c" would be 9.
Let's put them into the equation now.
[tex](x+5)^2 + (x-3)^2 = 9^2[/tex]
Simplify:
[tex](x^2 + 10x + 25) + (x^2-6x+9) = 81[/tex]
[tex]2x^2 + 4x + 34 = 81[/tex]
[tex]2(x^2+2x+17) = 81\\x^2+2x+17 = \frac{81}{2}[/tex]
[tex]x^2 +2x+ (17-\frac{81}{2})=0\\ x^2+2x+(\frac{-47}{2}) = 0\\[/tex]
From there, you can solve to x (it might be a little tricky, so sorry if you don't understand these steps).
[tex]2x^2 + 4x - 47 = 0\\\frac{-b ±(\sqrt{b^2-4ac})}{2a}[/tex]
final answer: [tex]x = \frac{-2+7\sqrt{2} }{2}[/tex] and [tex]\frac{-2-7\sqrt{2} }{2}[/tex] >
Hope this helped. sorry for the bad explanation but feel free to ask anything else.
