Find the value of $k$ in this question

Question

Suppose $x_1$ and $x_2$ are solutions to $x^2+x+k=0$, if $x_1^2+x_1x_2+x_2^2=2k^2$, find the value of $k$.

$x_1+x_2 = -1$
$x_1x_2 = k$

$(x_1+x_2)^2 = x_1^2+2(x_1x_2)+x_2^2=2k^2$

$=\frac {x_1^2}{2}+x_1x_2+\frac {x_2^2}{2}=2k^2=\frac{1}{2}$

$2k^2=\frac{1}{2}$
$k^2=\frac{1}{4}$
$k=\sqrt{\frac{1}{4}}$

Could someone please confirm if this is correct and perhaps identify any errors I may have made?

Answer 1

You are wrong in this line $~(x_1+x_2)^2 = x_1^2+2(x_1x_2)+x_2^2\color{red}{=2k^2}~$. The value in the left hand side is not equal to $~{2k^2}~$ due to the present of $~\bf 2~$ in the middle term. Here is the complete solution (as per the process you follow).

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$$(x_1+x_2)^2 = x_1^2+2x_1x_2 +x_2^2$$ $$\implies (-1)^2=(x_1^2+x_1x_2 +x_2^2)+x_1x_2$$ $$\implies 1=2k^2+k$$ $$\implies 2k^2+k-1=0$$ $$\implies k=\dfrac{-1\pm \sqrt{1^2-4\cdot 2\cdot(-1)}}{4}$$ $$\implies k=\dfrac{-1\pm 3}{4}$$ $$\implies k=-1,~~\dfrac 12$$

Answer 2

$$(x_1^2+x_2^2+x_1x_2)=(x_1+x_2)^2-x_1x_2=1-k=2k^2\implies 2k^2+k-1 \implies k=-1,1/2$$