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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Updated on: 10-Oct-2022

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