Question

Find the roots of quadratic equation √2x^2-2x-√3

Answer 1

Answer:

√2x^2-2x-√3

D=(-2)^2-4*√2*(-√3)

D=4+4√6

x=2^2+-√4+4√6/2√2

x=4+-2√6/2√2

x=4+2√6/2√2or4-2√6/2√2

Answer 2

x = $\dfrac{1+\sqrt{1+\sqrt{6}}}{\sqrt{2}}$ or, $\dfrac{1-\sqrt{1+\sqrt{6}}}{\sqrt{2}}$

Step-by-step explanation:

The given quadratic equation:

$\sqrt{2}x^2 -2x -\sqrt{3}$

Here, a = $\sqrt{2}$ , b = - 2 and c = $-\sqrt{3}$

To find, the roots of the given quadratic equation = ?

∵ D = $b^{2}-4ac$

= $(-2)^{2}-4(\sqrt{2})(-\sqrt{3})$

= $4+4\sqrt{6}$

= $4(1+\sqrt{6})$ > 0, the roots are real and unequal.

∴ $x=\dfrac{-b±\sqrt{D}}{2a}$

$=\dfrac{-(-2)±\sqrt{4(1+\sqrt{6})}}{2\sqrt{2}}$

$=\dfrac{2±2\sqrt{1+\sqrt{6}}}{2\sqrt{2}}$

$=\dfrac{2(1±\sqrt{1+\sqrt{6}})}{2\sqrt{2}}$

$=\dfrac{1±\sqrt{1+\sqrt{6}}}{\sqrt{2}}$

= $\dfrac{1+\sqrt{1+\sqrt{6}}}{\sqrt{2}}$ or, $\dfrac{1-\sqrt{1+\sqrt{6}}}{\sqrt{2}}$

∴ $\dfrac{1+\sqrt{1+\sqrt{6}}}{\sqrt{2}}$ or, $\dfrac{1-\sqrt{1+\sqrt{6}}}{\sqrt{2}}$