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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Updated on: 10-Oct-2022

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