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If the equations $x^{2}+k x+64=0$ and $x^{2}-8 x+k=0$ have real roots, then what is the value of $k ?$
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The correct answer is:
16
Givenequations are $x^{2}+k x+64=0$ and $x^{2}-8 x+k=0$...(i)
Since both the eqns have real roots, discriminant $\geq 0$...(ii)
$=b^{2} \geq 4 a c$
from eq $^{\mathrm{n}}(\mathrm{i})$, we have
$k^{2} \geq 4(64) \Rightarrow k^{2} \geq 256$
$\Rightarrow k \geq 16$...(A)
and from $\mathrm{eq}^{\mathrm{n}}$ (ii), we have
$\Rightarrow \quad 64 \geq 4 k \Rightarrow 4 k \leq 64$
$\Rightarrow k \leq 16$$\ldots(\mathrm{B})$
Hence, from eq $^{\mathrm{n}}(\mathrm{A})$ and $(\mathrm{B})$, we have $\mathrm{k}=16$
Since both the eqns have real roots, discriminant $\geq 0$...(ii)
$=b^{2} \geq 4 a c$
from eq $^{\mathrm{n}}(\mathrm{i})$, we have
$k^{2} \geq 4(64) \Rightarrow k^{2} \geq 256$
$\Rightarrow k \geq 16$...(A)
and from $\mathrm{eq}^{\mathrm{n}}$ (ii), we have
$\Rightarrow \quad 64 \geq 4 k \Rightarrow 4 k \leq 64$
$\Rightarrow k \leq 16$$\ldots(\mathrm{B})$
Hence, from eq $^{\mathrm{n}}(\mathrm{A})$ and $(\mathrm{B})$, we have $\mathrm{k}=16$
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