Question

Can anyone help me solve

Answer 1

Answer:

$x = +2\sqrt{2}i , {\textrm{ or}} x= -2\sqrt{2}i$

Step-by-step explanation:

Here,the given expression is : $5 + x^{2} = 2x^{2} + 13$

Now simplifying the expression, we get

$5 + x^{2} = 2x^{2} + 13$

⇒  $x^{2} - 2x^{2} = 13 - 5$

or, $-x^{2} = 8$

⇒  $x^{2} = -8 \\ or, x^{2} + 8 = 0$

Now, Since here the roots are NOT REAL but imaginary.

So, try and solve this by $x = \frac{b \pm \sqrt{b^{2} - 4ac } }{2a}$

here, a = 1, b = 0 and c = 8

So, solving this, we get $x = \frac{\pm \sqrt{-32} }{2} = \frac{\pm 4\sqrt{2} }{2}$

⇒ $x = +2\sqrt{2}i , {\textrm{ or}} x= -2\sqrt{2}i$