Answer:
[tex]x = 3 + 4\sqrt{3}[/tex]
Step-by-step explanation:
OPQ is a right angle triangle.
Using Pythagoras
[tex]x^{2} + (x+5)^{2} = (x+8)^{2}\\x^{2} + x^{2} + 10x +25 = x^{2} + 16x + 64\\x^{2} + x^{2} - x^{2} + 10x - 16x + 25 - 64 = 0\\x^{2} - 6x -39 = 0[/tex]
Using quadratic formula
[tex]x = \frac{-b \± \sqrt{b^{2}-4ac}}{2a}\\x = \frac{-(-6) \± \sqrt{(-6)^{2}-4(1)(-39)}}{2(1)}\\x = \frac{6}{2} \± \frac{\sqrt{192}}{2}\\x = 3 \± \frac{8\sqrt{3}}{2}\\x = 3 \± 4\sqrt{3}[/tex]
[tex]x = 3 + 4\sqrt{3} = 9.928(4s.f.)[/tex]
OR
[tex]x = 3 - 4\sqrt{3} = -3.928(4s.f.)[/tex] (Length can't be negative)
∴ [tex]x = 3 + 4\sqrt{3}[/tex]
