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# Write all the values of k for which the quadratic equation $x^2+kx+16=0$ has equal roots. Find the roots of the equation so obtained.

Given:

Given quadratic equation is $x^2 + kx + 16 = 0$.

To do:

We have to find the value of k for which the given quadratic equation has equal roots.

Solution:

$x^2 + kx + 16 = 0$

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=1, b=k$ and $c=16$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

$D=(k)^2-4(1)(16)$

$D=k^2-64$

The given quadratic equation has equal roots if $D=0$.

Therefore,

$k^2-64=0$

$k^2-(8)^2=0$

$(k+8)(k-8)=0$

$k+8=0$ or $k-8=0$

$k=-8$ or $k=8$

The values of k are $-8$ and $8$.

For $k = -8$,

$x^2 + kx + 16 = 0$

$x^2 + (-8)x + 16 = 0$

$x^2 - 8x + 16 = 0$

$(x - 4)^2 = 0$

$x-4=0$

$x=4$

Therefore, for $k=-8$ the roots of the given quadratic equation are $4$ and $4$.

For $k = 8$,

$x^2 + kx + 16 = 0$

$x^2 + 8x + 16 = 0$

$(x + 4)^2 = 0$

$x+4=0$

$x=-4$

Therefore, for $k=8$ the roots of the given quadratic equation are $-4$ and $-4$.

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