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# inscribed triangle question - help needed!

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inscribed triangle question - help needed! [#permalink]

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26 Jul 2012, 12:16
Hi.

i had an inscribed angle problem asking for the lenght of one of the sides.

the answer key says I am supposed to get this by factoring this quadratic y^2 + 8Y + 8 = 0.

how do i do that?

thanks,

Clearmountain
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Re: inscribed triangle question - help needed! [#permalink]

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26 Jul 2012, 12:51
clearmountain wrote:
Hi.

i had an inscribed angle problem asking for the lenght of one of the sides.

the answer key says I am supposed to get this by factoring this quadratic y^2 + 8Y + 8 = 0.

how do i do that?

thanks,

Clearmountain

If the answer is $$4+2\sqrt{2}$$ then the equation must be $$y^2-8y+8=0$$.
You can use the formula which gives the roots of a quadratic equation in terms of its coefficient.

Another way is as follows:
$$y^2-8y+8=y^2-8y+16-8=(y-4)^2-8=0$$ from which $$(y-4)^2=8$$ and either $$y-4=2\sqrt{2}$$, so $$y=4+2\sqrt{2}$$,
or $$y-4=-2\sqrt{2}$$, which gives $$y = 4-2\sqrt{2}$$.
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Re: inscribed triangle question - help needed! [#permalink]

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28 Jul 2012, 11:14
clearmountain wrote:
Hi.

i had an inscribed angle problem asking for the lenght of one of the sides.

the answer key says I am supposed to get this by factoring this quadratic y^2 + 8Y + 8 = 0.

how do i do that?

thanks,

Clearmountain
Hi Clearmountain,

EvaJager is correct--something must have slipped when you copied this question. After all, y^2 and 8 are both positive. That means that no positive value for y can satisfy the equation, because the sum of three positives will always be greater than 0! Can you double-check the equation and solution that you've posted? I'll be happy to check your math once I've seen what we're really supposed to do.

Regards,
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Re: inscribed triangle question - help needed!   [#permalink] 28 Jul 2012, 11:14
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