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sleet_krkn [62]
3 years ago
6

Can anyone help me solve

Mathematics
1 answer:
Goryan [66]3 years ago
6 0

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

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