Find the values of k for which the following equations have real roots
$kx(x-3) + 9 = 0$

Given:

Given quadratic equation is $kx(x-3)+9=0$.

To do:

We have to find the values of k for which the roots are real.

Solution:

$kx(x-3)+9=0$

$kx^2-3kx+9=0$

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

$a=k, b=-3k$ and $c=9$.

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

$D=(-3k)^2-4(k)(9)$

$D=9k^2-36k$

The given quadratic equation has real roots if $D≥0$.

Therefore,

$9k^2-36k≥0$

$9k(k-4)≥0$

$9k≥0$ and $k-4≥0$

$k≥0$ and $k≥4$

This implies,

$k≥4$

The value of k is greater than or equal to $4$.

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Updated on: 10-Oct-2022

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