Stumper

[barson90]

Determine all values of K such that twice the sum of the roots of Kx^2 -- 5x -- K + 7x^2 = 0 will be one more than four times the product of the roots.

 

[tkhunny]

Of only one could discern what is wanted. Please try again and make sure at least YOU can understand what you have written.

 

[soroban]

Hello, barson90!

On this planet, we do not use double minus-signs.
I'll take a guess at what you meant . . .

$\displaystyle \text{Determine all values of }K\text{ such that twice the sum of the roots of: }\:K\!x^2 \:-\: 5x \:-\: K +\: 7x^2 \;=\; 0$

$\displaystyle \text{will be one more than four times the product of the roots.}$

\(\displaystyle \text{We have: }\K+7)x^2 - 5x - K \:=\:0 \quad\Rightarrow\quad x^2 - \frac{5}{k+7}x - \frac{K}{K+7} \:=\:0\)

$\displaystyle \text{The sum of the roots is: }\:S \:=\:\frac{5}{K+7}$
\(\displaystyle \text{The product of the roots is: }\ \:=\:-\frac{K}{K+7}\)


$\displaystyle \underbrace{\text{"Twice the product }}_{2P} \underbrace{\text{ will be }}_{=} \underbrace{ \text{ four times the sum }}_{4S} \underbrace{\text{ plus one" }}_{+ 1}$

$\displaystyle \text{We have: }\:2\left(-\frac{K}{K+7}\right) \;=\;4\left(\frac{5}{K+7}\right) + 1 \quad\Rightarrow\quad \frac{-2K}{K+7} \;=\;\frac{20}{K+7} + 1$


$\displaystyle \text{Multiply by }(K+7)\!:\;\;-2K \;=\;20 + K + 7 \quad\Rightarrow\quad -3K \:=\:27$

. . $\displaystyle \text{Therefore: }\;\boxed{K \:=\:-9}$


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I checked this result and it works out.

But the algebra is a Herculean labor.
. . The two roots are complex.