Question

Find the root of
$\sqrt{2x { }^{2} } + x + \sqrt{2} = 0$
​

Answer 1

Answer:

The root of    x = -2 + √2

Step-by-step explanation:

Step(i):-

Given $\sqrt{2x^{2} } + x +\sqrt{2} = 0$

      ⇒    $\sqrt{2} } ( x^2 ) ^{\frac{1}{2} } + x +\sqrt{2} = 0$

      ⇒   $\sqrt{2}x + x +\sqrt{2} = 0$

     ⇒      $(\sqrt{2} + 1) x = - \sqrt{2}$

           $x = \frac{-\sqrt{2} }{(\sqrt{2} + 1)}$

Step(ii):-

Rationalizing

     $x = \frac{-\sqrt{2} }{(\sqrt{2} + 1)} X \frac{(\sqrt{2} - 1)}{(\sqrt{2} - 1)}$

Apply ( a-b ) ( a+b) = a²-b²

    $x = \frac{-\sqrt{2}(\sqrt{2} -1) }{(\sqrt{2})^{2} - 1^{2} )} = \frac{-(2-\sqrt{2} }{2-1}$

The root of    x = -2 + √2