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How many $\underline{\text{negative roots}}$ does the equation $x^4-5x^3-4x^2-7x+4=0$ have?

My reasoning:

I rewrote the equation like: $$x^4-5x^3-4x^2-7x+4=0 \Rightarrow (x^2-2)^2 = 5x^3+7x$$

For any negative $x$, the outcome is never a negative term in the left member of the equation, and always a negative term in the right member of the equation. That's impossible, so the answer is that there are no negative roots for the original equation.

Does anyone see another way out?

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You mean $(x^2 - 2)^2 = 5x^3 + 7x$? –  Deven Ware Jan 29 '13 at 19:41
O yeah, my bad. Thx. –  Sawyier Jan 29 '13 at 19:42
Either way, you are correct, there are no negative roots. –  Deven Ware Jan 29 '13 at 19:43
Are yu familiar with Descartes' rule of signs? Regards –  Amzoti Jan 29 '13 at 19:45
For any negative $x$ tested? Can't you see immediately that $(x^2 - 2)^2 \ge 0$ for all real $x$, while $5 x^3 + 7 x < 0$ for all negative $x$? –  Robert Israel Jan 29 '13 at 19:47

1 Answer 1

up vote 1 down vote accepted

You might want to say explicitly that $\forall x < 0,\; x\neq -\sqrt{2}$, the left-hand side of the equation is positive (whatever the value of $x$ - save for $x = \pm \sqrt{2}$, in which case the LHS $= 0$), while the right hand side of the equation is negative $\forall x<0$, since $5x^3 + 7x = x(x^2 + 7)$, evaluates to the sign of $x$. This is impossible for any equation. Hence, there is no solution (i.e., root) $x$ such that $x \lt 0$.

Essentially, what you argued :-)

Certainly, there is no need to find the actual roots, as you observe.

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