Question

If the roots of x²+kx+k+3=0 are real and equal, what is the value of k?​

Answer 1

Answer:

Given :-

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

To Find :-

• What is the value of k.

Formula Used :-

$\clubsuit$ Discriminant Formula :

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

Solution :-

Given Equation :

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

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

✫ a = 1

✫ b = k

✫ c = k + 3

According to the question by using the formula we get,

In the question, it is given that the roots are real and equal. So,

$\implies \sf\bold{\purple{b^2 - 4ac =\: 0}}$

By putting all the values we get,

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

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

$\implies \sf k^2 - 4k - 12 =\: 0$

By doing middle term break we get,

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

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

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

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

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

$\implies \sf\bold{\red{k =\: 6}}$

Or,

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

$\implies \sf\bold{\red{k =\: - 2}}$

$\therefore$ The value of k is 6 and - 2 .

Answer 2

Answer:

The value of k is 6 or - 2

Step-by-step explanation:

Given equation is

$x^{2}$ + kx + (k + 3) = o

The general formula of quadratic equation is;

a$x^{2}$ + bx +c =0

By comparing two equations we get,

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

We know that,

When roots are real D =0

D = $b^{2}$ -4ac

Putting the value of D,a,b,c

0 = $k^{2}$ - 4 ×1 × ( k + 3)

$k^{2}$ = 4 ( k +3)

$k^{2}$ = 4k + 12

$k^{2}$-  4k - 12 =0

$k^{2}$ - 6k + 2k -12 = 0     (by middle term factorization)

k ( k - 6 ) +2 ( k - 6) = 0

(k - 6 ) ( k + 2 ) = 0

Either (k-6 )= 0    or  (k +2) =0

           k = 6   or    k = -2