Question

find he value of k for which the quadratic equation is (k+4)x2 +(k+1)x +1=0 has equal roots.

Answer 1

Answer:

The value of k is either -3 or 5.

Step-by-step explanation:

If a quadratic equation $ax^2+bx+c=0$ has two equal root, then

$b^2-4ac=0$.

The given quadratic equation is

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

Here,

$a=k+4, b=k+1,c=1$

It is given that the given equation has two equal roots. So,

$b^2-4ac=0$

$(k+1)^2-4(k+4)(1)=0$

$k^2+2k+1-4k-16=0$

$k^2-2k-15=0$

Splitting the middle term, we get

$k^2-5k+3k-15=0$

$k(k-5)+3(k-5)=0$

$(k+3)(k-5)=0$

Using zero product property, we get

$k+3=0\Rightarrow k=-3$

$k-5=0\Rightarrow k=5$

Therefore, the value of k is either -3 or 5.

Answer 2

Answer:

The answer to the question you have asked is here.....................OK!