Using Pythagoras Theorem, find the value of $x$ in this triangle.

Question

Using Pythagoras Theorem [Baudhāyana Śulbasûtra], find the value of $x$ in this triangle.

I tried to solve it but I seem to get stuck with end result of $$26x+5x^2=588$$

Answer 1

Your equation is not corret: im fact you forgot some squares as for example $25$ that becomes $625$.

By Pythagoras's theorem, we have;: $$16(x+2)^2+(x+3)^2=625\leftrightarrow 17x^2+70x-552=0$$ The solutions are: $$x=\frac{-70\pm\sqrt{70^2+4\cdot17\cdot552}}{34}=\frac{-70\pm206}{34}=-\frac{138}{17} \vee 4$$ The first solutions can't be accepted, while the second is correct.

Answer 2

Firstly to solve your equation, rearrange to $$5x^2+26x-588=0$$

$$x=\frac{-26\pm\sqrt{26^2-4(5)(588)}}{2(5)}$$ as given by the quadratic formula. (This cannot be factorised.)

however, this is not correct. Looking at your question, it seems that you forgot to also square the $4$. Remember that $(2x)^2=4x^2$ not $2x^2$. Try to fix it, then solve the quadratic equation.

Cheers! :)