Question

.Find the value of k for which the quadratic equation (k + 4) x2 + (k+1) x + 1 = 0 has equal
roots.
$2 {}^{2}$
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Answer 1

Given:

A quadratic equation

$(k+4)x^{2} + (k+1)x +1=0$

Find What :

If the roots of the equation are equal then

find the value of k

Solution:

We know that if the roots of the equation

$ax^{2} + bx +c = 0$

are equal then

$b^2=4ac-------------------------(1)$

In the given equation

a=(k+4) ,  b= k+1 and c =1

Putting the values in (1) we get

$(k+1)^2 =4(k+4) \times 1\\\\k^2 +2k+1=4k+4\\\\k^2 -2k-3=0\\\\k^2-3k+k-3=0\\\\k(k-3)+(k-3)=0\\\\(k-3)(k+1)\\\\k-3=0,k=3\\\\OR\\\\k+1=0,k=-1\\\\Thus\, k=3\,\, or -1$

Answers:

k = 3 or -1