172
Find the value(s) of $k$ for which the equation $2x^2 + kx + 3 = 0$ has real and equal roots. Hence, find the roots of the equations so obtained.
Show SolutionHide Solution↓
For equal roots; $b^2 - 4ac = 0$
$k^2 - 24 = 0$
$\Rightarrow k = \pm 2\sqrt{6}$
Equations are
$2x^2 + 2\sqrt{6}x + 3 = 0$; $2x^2 - 2\sqrt{6}x + 3 = 0$
Roots are $x = -\sqrt{\frac{3}{2}}, -\sqrt{\frac{3}{2}}$; $x = \sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}}$
$k^2 - 24 = 0$
$\Rightarrow k = \pm 2\sqrt{6}$
Equations are
$2x^2 + 2\sqrt{6}x + 3 = 0$; $2x^2 - 2\sqrt{6}x + 3 = 0$
Roots are $x = -\sqrt{\frac{3}{2}}, -\sqrt{\frac{3}{2}}$; $x = \sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}}$