Question

find the root of the quadratic eqation = x^2-4x-8=0​

Answer 1

Answer:

$x=2+2\sqrt{3}$ or $2-2\sqrt{3}$

Explanation:

$x^{2} -4x-8=0\\Here, a=1, b=-4, c=-8\\So, D = b^{2} -4ac=(-4)^{2}-4(1)(-8)\\D=16+32\\D=48\\\sqrt{D} = \sqrt{48} = 4\sqrt{3} \\Therefore, x=\frac{-b+\sqrt{D} }{2a} \\x= \frac{-(-4)+4\sqrt{3}}{2} =2+2\sqrt{3} \\\\OR\\x=\frac{-b-\sqrt{D} }{2a} \\x= \frac{-(-4)-4\sqrt{3}}{2} =2-2\sqrt{3}$

Hence, $x=2+2\sqrt{3}$ or $2-2\sqrt{3}$