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By completing the square find in terms of $k$ the roots of the equation $$x^2 + 2kx-7=0$$ prove for all real values of $k$, the roots are real

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  • $\begingroup$ $(x+k)^2 -k^2-7=0$. $\endgroup$ – njguliyev Oct 6 '13 at 15:11
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$$x^2+2kx-7=x^2+2\cdot x\cdot k+k^2-k^2-7=0$$ $$(x+k)^2=k^2+7$$ $$x+k=\pm\sqrt{k^2+7}$$ $$x=-k\pm\sqrt{k^2+7}$$ because $k^2+7\geq 0$ for all real $k$ all roots are real.

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