Question

find the value of k for which the quadratic equation 2x^2+kx+3=0 has two real equal roots​

Answer 1

For real and equal roots :-

$D=0$

Where,

D is the discriminant of the quadratic equation.

D=b²-4ac

Hence,

P(x) = 2x² + kx +3

Thus,

Discriminant of the given equation is:-

D =b² - 4ac

D = k² - 4(2)(3)

D =k² - 24

We have to equate the with zero

Therefore :-

${k}^{2} - 4(3)(2) = 0 \\ \\ \\ {k}^{2} - 24 = 0 \\ \\ \\k = \pm\sqrt{24} \\ \\ \\k = \pm2 \sqrt{6}$

Thus for these values of k, the quadratic equation can have real and equal roots.

Answer!

Answer 2

Answer:

$k=±2 \sqrt{6}$

Step-by-step explanation:

$(D=0)</p><p>$

D=b²-4ac

P*x = 2x² + kx +3

D =b² - 4ac

D = k² - 4

D =k² - 24

$\begin{lgathered}{k}^{2} - 4 = 0 \\ {k}^{2} - 24 = 0 \\k = \pm\sqrt{24} \\k = \pm2 \sqrt{6}\end{lgathered}$