Question

The values of k for which the
quadratic equation
${x}^{2}$
- kx + k + 3 =
0 has coincident roots, are​

Answer 1

Answer :-

If the equation has coincident roots, then the value of discriminant is equal to zero.

→ $\sf D = b^2 - 4ac$

• $\sf a = 1$
• $\sf b = -k$
• $\sf c = k + 3$

→ $\sf b^2 - 4ac = 0$

→ $\sf (-k)^2 - 4 \times 1 \times ( k + 3 ) = 0$

→ $\sf k^2 - 4(k + 3) = 0$

→ $\sf k^2 - 4k - 12 = 0$

→ $\sf k^2 - 6k + 2k - 12 = 0$

→ $\sf k( k - 6) + 2( k - 6) = 0$

→ $\sf (k + 2)( k - 6) = 0$

Either ( k + 2 ) is 0 or ( k - 6 ) is 0.

→ $\sf k + 2 = 0$

• $\sf k = -2$

→ $\sf k - 6 = 0$

• $\sf k = 6$

Values of k = -2 , 6

Answer 2

QuestioN :

The values of k for which the quadratic equation  x² - kx + k + 3 =  0 has coincident roots, are​

GiveN :

equation  x² - kx + k + 3 =  0

To FiNd :

The values of k

ANswer :

The values of k is -2 , 6

SolutioN :

$\sf \rightarrow U = s^2-4rt$

$\sf Here,$

• $\sf r = 1$

• $\sf s = -k$

• $\sf t = k+3$

$\sf now \:solving \:the\: equation,$

$\sf \rightarrow s^2 -4rt=0$

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

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

$\sf \rightarrow k^2 -4k-12 = 0$

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

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

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

$\sf Either\: ( k + 2 )\: is\: 0 \:or\: ( k - 6 )\: is\: 0.$

$\sf The\: values\: of\: k\: is\: -2 , 6.$

∴Hence, The values of k is -2 , 6.

If you are not able to see the answer properly kindly view the answer from website by the given link.↓↓

brainly.in/question/37346352