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25 votes
25 votes
25x^-4-99x^-2-4=0
How do I solve for this?

User Mahatma Aladdin
by
2.7k points

1 Answer

5 votes
5 votes

Given


25x^(-4) - 99x^(-2) - 4 = 0

consider substituting
y=x^(-2) to get a proper quadratic equation,


25y^2 - 99y - 4 = 0

Solve for
y ; we can factorize to get


(25y + 1) (y - 4) = 0


25y+1 = 0 \text{ or } y - 4 = 0


y = -\frac1{25} \text{ or }y = 4

Solve for
x :


x^(-2) = -\frac1{25} \text{ or }x^(-2) = 4

The first equation has no real solution, since
x^(-2) = \frac1{x^2} > 0 for all non-zero
x. Proceeding with the second equation, we get


x^(-2) = 4 \implies x^2 = \frac14 \implies x = \pm√(\frac14) = \boxed{\pm \frac12}

If we want to find all complex solutions, we take
i=√(-1) so that the first equation above would have led us to


x^(-2) = -\frac1{25} \implies x^2 = -25 \implies x = \pm√(-25) = \pm5i

User CoffeJunky
by
2.8k points