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I have right angled triangle I've been attempting to prove a quadratic equation with for a while.

It has a hypotenuse of $2x + 1 cm$, a base of $x + 5 cm$, and height of $x - 2 cm$. I calculated its area to be $x^2 + 3x - 10$, but am now confused.

I am attempting to use this triangle to show that $x^2 - x - 14 = 0$. Could anyone prod me in the right direction?

GregW
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1 Answers1

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Using Pythagoras' theorem you have:

$$ \begin{split} && (2x+1)^2=(x+5)^2+(x-2)^2 \\ &\implies& 4x^2+4x+1=x^2+10x+25+x^2-4x+4 \\ &\implies& x^2-x-14=0 \end{split} $$

Sam Weatherhog
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