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If the equation $$x^8 – 3abx + a^2 = 0$$ has one root that is $\ge 3$. And if $a\geq 0$ .

Can $b$ be equal to 15?

I took One root of this equation is (3 + t), where t ≥ 0 And put the value of this root in equation.

Thanks in advance.

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$a$ is what?... – Daniel Montealegre Mar 13 '12 at 8:23
@Daniel it is not given .It is positive. – vikiiii Mar 13 '12 at 8:26
Think of it as a quadratic in $a$. – Brian M. Scott Mar 13 '12 at 8:51
@Brian M. Scott Actually it is a multiple choice question. I know the answer is cannot be equal to 15.Thats why i have made up my question according to that.There are 4 more options. – vikiiii Mar 13 '12 at 9:08
Does x belong to the set of real numbers? – Tomarinator Mar 18 '12 at 15:09
up vote 2 down vote accepted

Hint: Try to solve for $a$ in terms of $x$ when $b=15$

Your solution should include a square root. Can you spot anything about this square root when $x \ge 3$?

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