Determine the set of values of k for which the following  quadratic equations have real roots:
$4x^2-3kx+1=0$

Given:

Given quadratic equation is $4x^2 - 3kx + 1 = 0$.

To do:

We have to find the value of k for which the given quadratic equation has real roots.

Solution:

$4x^2 - 3kx + 1 = 0$

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

$a=4, b=-3k$ and $c=1$.

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

$D=(-3k)^2-4(4)(1)$

$D=9k^2-16$

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

Therefore,

$9k^2-16≥0$

$(3k)^2-(4)^2≥0$

$(3k+4)(3k-4)≥0$

$3k≤-4$ or $3k≥4$

$k≤\frac{-4}{3}$ or $k≥\frac{4}{3}$

The values of k are $k≤\frac{-4}{3}$ and $k≥\frac{4}{3}$.

Tutorialspoint

Updated on: 10-Oct-2022

30 Views

Kickstart Your Career

Get certified by completing the course

Get Started