Question

find an upper and lower limit for the zero of the following function f(x)=x^4-4x^3+x^2+6x+2

Answer 1

you mean the positive and negative roots?

Answer 2

yes

Answer 3

\[A = 1, B =-4, C = 1, D =6, E = 2\]
\[\text{let }x = u - \frac{B}{4A} = u+1\]
\[(u+1)^4 -4(u+1)^3 + (u+1)^2+6u+8 = 0\]
\[\text{expand the binomial coefficients}\]
\[(u^4 + 4u^3++u^2 + 4u + 1)-4(u^3+3u^2+3u+1)+(u^2+2u+1)+6u+8=0\]

Answer 4

mistake in the last line
\[(u^4 + 4u^3+6u^2 + 4u + 1)-4(u^3+3u^2+3u+1)+(u^2+2u+1)+6u+8=0\]

Answer 5

collect the coefficients of the same powers of u,\[u^4 -5u^2+6 = 0\]

Answer 6

\[x^4-4 x^3+x^2+6 x+2=\left(x^2-2 x-2\right) \left(x^2-2 x-1\right) \]The zeros are:\[\left\{1-\sqrt{3},1-\sqrt{2},1+\sqrt{2},1+\sqrt{3}\right\} \]\[\{-0.732051,-0.414214,2.41421,2.73205\} \]A plot of the expression is attached.

Answer 7

wow, how did you factor it so easily

Answer 8

I'm still going to have to substitute u^2 for p and solve \[p^2 - 5p + 6 = 0\]

Answer 9

I cheated and used Mathematica. Selected the expression with the Mouse and clicked on the "Factor" button.

Answer 10

:( Wish I owned mathematica

Answer 11

well for \[p^2 - 5p + 6 = 0\]you can complete the square, use the quadratic formula for the roots, or follow your heart to easily find that the roots are \[p = 2 \text{ or } p = 3\]

Answer 12

They practically give it away for Math majors. I paid $250 US for Mathematica 8 Home Edition. It's a large system. The installation file is around 500Mb.

Answer 13

\[p = u^2\]so\[u = \sqrt{p}\ \ or\ \ -\sqrt{p} \]

Answer 14

\[u = \sqrt{2}\ \ or\ \ -\sqrt{2}\]or\[u = \sqrt{3}\ \ or\ \ -\sqrt{3}\]

Answer 15

and \[x = u+1\]so\[x_1 = \sqrt{2}+1\]
\[x_2 = -\sqrt{2}+1\]
\[x_3 = \sqrt{3} +1\]
\[x_4 = -\sqrt{3} + 1\]

Answer 16

all that trouble for a quartic function doesn't seem worth it... I think there's an easier way

Answer 17

the upper zero of the function would be the larger of the roots, I think, which is\[x_3 = \sqrt{3} + 1\]
the lower limit would be the smallest of the roots, which is \[x_4 = - \sqrt{3}+1\]