Question

For what values of k, the roots of the quadratic equation (k+4)x2+(k+1)x+1 = 0 are equal?
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Answer 1

${} \huge \underline \red{Question - }$

For what values of k, the roots of the quadratic equation (k+4)x2+(k+1)x+1 = 0 are equal?

${} \huge \underline \blue{Solution - }$

The given Quadratic Equation is-

(k+4)x2+(k+1)x+1 = 0

For Equal Roots, D should be equal to zero.

Discriminant D of the quadratic equation ax2+bx+c = 0 is given by-

[ D = b2-4ac ]

Comparing the equation ax2+bx+c = 0 with given quadratic equation (k+4)x2+(k+1)x+1 = 0, we get-

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

∴ D = 0

⇒ (k+1)2-4(k+4) = 0

⇒ k2+2k+1-4k-16 = 0

⇒ k2-2k-15 = 0

⇒ k2-5k+3k-15 = 0

⇒ k(k-5)+3(k-5) = 0

⇒ (k+3)(k-5) = 0

∴ k=-3 or k=5

Hence, the possible values of k are -3 and 5.

Answer 2

Answer:

The value of k is either -3 or 5.

Step-by-step explanation:

If a quadratic equation has two equal root, then

.

The given quadratic equation is

Here,

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

Splitting the middle term, we get

Using zero product property, we get

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

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