x^2-38x+326=0 please factor

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- anonymous

x^2-38x+326=0 please factor

- katieb

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- anonymous

\[x=19\pm \sqrt{35}\]

- anonymous

Just apply the quadratic formula

- anonymous

use the quadratic formula
\[x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}, \]
which
A is the coefficient for the x^2 term
B is the coefficient for the x term
C is the coefficient for the x^0 term

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## More answers

- anonymous

Yup, robtotey has the correct answer and Kynosis gives the formula

- anonymous

Well it looks like I will need to start this from the beginning. This is new to me so I am lost. I have a word problem. It is about a ladder leaning against a bldg. distance from bottom of ladder to bldg is 18' less than the length of the ladder. How high up the side of the bldg is the top of the ladder if that distance is 1' less than the length of the ladder

- anonymous

Maybe I set it up wrong. I started with \[(x-18)^2+(x-1)^2=x^2\]

- anonymous

Is this right?

- anonymous

I agree with you Geminiman

- anonymous

finding the roots give you the factor of the equation indirectly. because equation are in the quadratic form is (x - n) ( x- n) with n being the root you have a factored equation. of \[( x - 19+√(35)) ( x - 19-√(35))\]

- anonymous

ok. then I did this formula: \[x^2-36+324+x^2-2x+2=x^2\]

- anonymous

I get x^2 - 38x + 325 = 0

- anonymous

ok I got x^2338x+326=0

- anonymous

Then you get (x - 25)(x - 13) = 0 which is
x = 25, 13

- anonymous

sorry typed it wrong. x^2-38x+326=0

- anonymous

When you took (x - 1)^2 Gem you had x^2 - 2x + 2 and it is actually x^2 - 2x + 1

- anonymous

ah ok that is why I got 326 instead of 325.

- anonymous

Thank you all for your help

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