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I have this equation $f(x) = \frac{1}{4}x^4-2x^3+6x^2-13x+4$

I am asked to calculate by hand the one real root out of it. When derivated $x^3-6x^2+12x-13=0$

This is where I am stuck at. I know the root is $\sqrt[3]{5}+2$.

I have been trying to calculate the answer in many different ways and still not getting the correct answer. Using common divisor etc. What am I doing wrong?

f1tz
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1 Answers1

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$x^3-6x^2+12x-13= (x - 2)^3 -5=0$

Therefore, the only real root is $\sqrt[3]{5}+2$

Basically, you had to complete the "cube" which is a pretty standard technique.

Why are you reading my name
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  • Uh, this does not answer the question, which asks to compute the real roots of the quartic. You have found the real minima, it still needs to be shown that this evaluates to zero in the quartic. But this does not occur, hence 2 real roots, as $f < 0$ at this point. – Mummy the turkey Feb 06 '21 at 12:15