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# algebra

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Let a and b be the roots of kx^2 + 4x + 4 = 0.  If a^2 + b^2 = 24, then find all possible values of k.

Jul 7, 2020

#1
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k={{b,0}}

Jul 7, 2020
edited by whymenotsmart  Jul 7, 2020
#2
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By Vieta's formula,

$$a + b = -\dfrac4k\\ ab = \dfrac4k$$

We start with an identity to find the value of a2 + b2.

$$a^2 + 2ab + b^2 = (a + b)^2\\ a^2 + b^2 + 2\left(\dfrac4k\right) = \left(-\dfrac4k\right)^2\\ a^2 + b^2 = \dfrac{16}{k^2} - \dfrac8k =\dfrac{8(2 - k)}{k^2}$$

Then, we equate it to 24.

$$\dfrac{8(2-k)}{k^2} = 24\\ 3k^2 = 2 - k\\ 3k^2 + k - 2 = 0$$

I believe you can do the rest.

Jul 7, 2020