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If the equation $x^{2}+k^{2}=2(k+1) x$ has equal roots, then what is the value of $\mathrm{k} ?$
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Verified Answer
The correct answer is:
$-\frac{1}{2}$
$x^{2}+k^{2}=2(k+1) x$
$\Rightarrow x^{2}-2(k+1) x+k^{2}=0$
For roots to beequal discriminant $=0$ So, $\{-2(k+1)\}^{2}-4 k^{2}=0$
or $4(k+1)^{2}-4 k^{2}=0$
or $(\mathrm{k}+1)^{2}-\mathrm{k}^{2}=0$
or, $2 \mathrm{k}+1=0$
$k=-\frac{1}{2}$
$\Rightarrow x^{2}-2(k+1) x+k^{2}=0$
For roots to beequal discriminant $=0$ So, $\{-2(k+1)\}^{2}-4 k^{2}=0$
or $4(k+1)^{2}-4 k^{2}=0$
or $(\mathrm{k}+1)^{2}-\mathrm{k}^{2}=0$
or, $2 \mathrm{k}+1=0$
$k=-\frac{1}{2}$
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