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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 $k≥\frac{4}{3}$.