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Determine the set of values of k for which the following quadratic equations have real roots:
$x^2-kx+9=0$
Given:
Given quadratic equation is $x^2 - kx + 9 = 0$.
To do:
We have to find the value of k for which the given quadratic equation has real roots.
Solution:
$x^2 - kx + 9 = 0$
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 $c=9$.
The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.
$D=(-k)^2-4(1)(9)$
$D=k^2-36$
The given quadratic equation has real roots if $D≥0$.
Therefore,
$k^2-36≥0$
$k^2-(6)^2≥0$
$(k+6)(k-6)≥0$
$k≤-6$ or $k≥6$
The values of k are $k≤-6$ and $k≥6$.
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