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real solutions the equation is x^4+8x^2=9 could i have real solutions and complex...

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cleokay | Student, Undergraduate | (Level 1) eNoter

Posted January 11, 2011 at 9:32 AM via web

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real solutions

 the equation is x^4+8x^2=9

could i have real solutions and complex solutions for an equation?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted February 21, 2013 at 6:41 AM (Answer #3)

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You need to solve for x the equation `x^4+8x^2=9` , hence, you should complete the square to the left side, using the formula `(a+b)^2 = a^2 + 2ab + b^2,` such that:

`x^4 + 8x^2 + 16 = 9 + 16` (add `4^2 = 16` both sides to preserve the equation)

`(x^2 + 4)^2 = 25 => x^2 + 4 = +-sqrt(25) => x^2 + 4 = +-5`

`x^2 = 5 - 4 => x^2 = 1 => x_(1,2) = +-1`

 

`x^2 = -5 - 4 => x^2 = -9 => x_(3,4) = +-sqrt(-9) => x_(3,4) = +-3i`

Hence, evaluating the equation ` x^4+8x^2=9` yields that there exists four roots, two of them being complex, such that `x_(1,2) = +-1` and `x_(3,4) = +-3i.`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted January 12, 2011 at 10:17 AM (Answer #2)

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Of course, you can have both types of solutions for an equation.

In this case, the total number of solutions is 4.

The method of solving is to substitute x^2 = t, since we don't know to solve an equation of 4th order.

t^2 + 8t - 9  =0

We'll apply quadratic formula:

t1 = [-8+sqrt(64 + 36)]/2

t1 = (-8+10)/2

t1 = 1

t2 = -9

But x^2 = t => x^2 = 1 => x1 = 1 and x2 = -1

x^2 = -9

x3 = 3i

x4 = -3i

As we can see, the solutions x1 and x2 are real numbers and the solutions x3 and x4 are imaginary. 

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