In the following, determine the set of values of k for which the given quadratic equation has real roots:
$2x^2 + 3x + k = 0$

Given:

Given quadratic equation is $2x^2 + 3x + k = 0$.

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,

$a=2, b=3$ and $c=k$.

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

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

$D=9-8k$

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

Therefore,

$9-8k≥0$

$9≥8k$

$k≤\frac{9}{8}$

Therefore, $k≤\frac{9}{8}$.

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Updated on: 10-Oct-2022

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