In the following, determine the set of values of k for which the given quadratic equation has real roots:
$kx^2 + 6x + 1 = 0$

Given:

Given quadratic equation is $kx^2 + 6x + 1 = 0$.

To do:

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

Solution:

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

$a=k, b=6$ and $c=1$.

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

$D=(6)^2-4(k)(1)$

$D=36-4k$

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

Therefore,

$36-4k≥0$

$36≥4k$

$k≤\frac{36}{4}$

$k≤9$

Therefore, $k≤9$.

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Updated on: 10-Oct-2022

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