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Find the values of k for which the given quadratic equation has real and distinct 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 given quadratic equation has real and distinct roots.
Solution:
$kx^2+6x+1=0$
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 and distinct roots if $D>0$.
Therefore,
$36-4k>0$
$36>4k$
$k<\frac{36}{4}$
$k<9$
The value of k is less than $9$.
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