Question

$\large\underline{\tt \widetilde{Question}}$

If the roots of x² + kx +k+3 = 0 are real and equal, what is the value of k?
(A) - 6 or 2
(B) 6 or-2
(C) 4 or -2
(D) – 4 or 2 - ​

Answer 1

Answer:

Given :-

• The roots of x² + kx + k + 3 = 0 are real and equal.

To Find :-

• What is the value of k.

Solution :-

Given Equation :

$\mapsto \bf x^2 + kx + k + 3 =\: 0$

By comparing with ax² + bx + c = 0 we get,

where,

❒ a = 1

❒ b = k

❒ c = k + 3

$\bigstar$ The roots are real and equal.

So,

$\bigstar \: \: \sf\boxed{\bold{Discriminant =\: b^2 - 4ac =\: 0}}\: \: \: \bigstar$

According to the question by using the formula we get,

$\implies \bf b^2 - 4ac =\: 0$

$\implies \sf (k)^2 - 4(1)(k + 3) =\: 0$

$\implies \sf k^2 - 4(k + 3) =\: 0$

$\implies \sf k^2 - 4k - 12 =\: 0\: \: \bigg\lgroup \bf By\: Doing\: Middle-term\: break \bigg\rgroup$

$\implies \sf k^2 - (6 - 2)k - 12 =\: 0$

$\implies \sf k^2 - 6k + 2k - 12 =\: 0$

$\implies \sf k(k - 6) + 2(k - 6) =\: 0$

$\implies \sf (k - 6)(k + 2) =\: 0$

$\implies \sf k - 6 =\: 0$

$\implies \bf k =\: 6$

Or,

$\implies \sf k + 2 =\: 0$

$\implies \bf k =\: - 2$

Hence, we get the two values of k i.e, 6 , - 2.

$\sf\bold{\underline{\therefore\: The\: value\: of\: k\: is\: 6\: or\: - 2\: .}}$

Hence, the correct options is option no (B) 6 or - 2 .

Answer 2

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The given quadratic equation is : x² + kx +k+3 = 0

Here,

a=1 , b=k , c=k+3

Now,

$\rm D=b^{2}-4×a×c$

$\rm = {k}^{2} - 4 \times 1 \times ( k+ 3)$

We know that a quadratic equation has real and equal roots If D=0

$\rm \implies {k}^{2} - 4k + 12 = 0$

$\rm\implies (k−6)(k+2)=0$

$\rm\implies When k-6=0 or k+2=0$

$\boxed{\color{purple} \rm\implies k=6,-2}$