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

`x = +2√(2)i ,  {\textrm{   or}} x= -2√(2)i`

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 = (\pm √(-32) )/(2) = (\pm 4√(2)  )/(2)`

⇒
`x = +2√(2)i ,  {\textrm{   or}} x= -2√(2)i`