Solving equations with degree higher than 2

Question

Please tell me what is wrong with the following procedure:

Q. Find solutions of the following expression.
 
(1⁄4)x^4 - (3⁄4)x^2 - 1 = 0

=> (1⁄4)x^4 - (3⁄4)x^2 = 1 -----------------moving 1 to RHS

=> (1⁄4)*(x^2) * (x^2 - 3) = 1 ---------- taking (1⁄4)*(x^2) common

=> (x^2)*(x^2 -3) = 4 ---------------- mulitplying 4 on both sides

=> x^2 = 4 OR x^2 - 3 = 4

=> x= +/-2 OR x = +/-7

Thus, there are four solutions: +2, -2, +7, -7.

However, the correct answer is that there are only two solutions: +2,-2.

KanishkM · Answer

Hey  jk11

you could have either substituted x^2 as some variable lets  say a , or you could have just used it as it is, said that, i took the  latter

x^2 will be +ive, x^4 will be positive

x^4 - 3x^2 - 4 = 0

x^2 (x^2 -4) + 1 (x^2-4) = 0

two solutions x^2 = -1 or  x^2 = 4 
only  this  is possible

x^2 = 4

two solutions -2 or 2

jk11 · Answer

I understand and agree with your approach. I just wanted to know what is wrong with mine? Which step made my solution wrong?

KanishkM · Answer

x^2 = 7 will be x = \sqrt{7} So the question would be asking for integer values There was nothing wrong in the approach, I prefer not working with fractions, more scope of error

Mandy001 · Answer

The problem arises after the step where you arrive

x^2((x^2)-3) = 4

After this, you take  x^2 = 4   or  (x^2)-3 = 4 
I don't know how you arrived at this step but let us go with it for once.

If  x^2 = 4  , then  (x^2)-3 = 4-3 = 1 
So therefore the product of both of the terms is 4*1=4.

Now if  (x^2)-3 = 4 , then  x^2 = 4+3 = 7  (using the same equation)
So therefore the product of the terms is 7*4=28. 
  Hence wrong

Solving equations with degree higher than 2

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