hi,

i need to solve x^2-18x+53=0 to find length of hypotenuse. i am unsure how to do this so some help would be greatly appreciated!

would i do quadratic equation on it??

thanks

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- Apr 6th 2011, 06:37 AMandyboy179find hypotenuse
hi,

i need to solve x^2-18x+53=0 to find length of hypotenuse. i am unsure how to do this so some help would be greatly appreciated!

would i do quadratic equation on it??

thanks - Apr 6th 2011, 07:24 AMearboth
This

$\displaystyle x^2-18x+53=0$

is a quadratic equation. To solve it for x use the quadratic formula:

The quadratic equation $\displaystyle ax^2 + bx +c = 0$ has the solution:

$\displaystyle x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Plug in the values of a, b and c which you already know. - Apr 6th 2011, 07:27 AMandyboy179
i can do that but i will give me two different answers for x however i only need one what do i do with both the answers?

- Apr 6th 2011, 07:28 AMTheChaz
Choose the positive answer, since the side length of a triangle can't be negative (up to obscure fields of study :) )

- Apr 6th 2011, 07:34 AMearboth