Question

Can anyone help me solve
5 + x^2= 2x^2+ 13
USING INVERSE OPERATIONS. and please provide the steps :)

OH, and please remember that you cannot take the square root of a negative number :)

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 $x = \frac{\pm \sqrt{-32} }{2} = \frac{\pm 4\sqrt{2}  }{2}$

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