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 March 16th, 2012, 04:28 PM #1 Member   Joined: Dec 2011 Posts: 44 Thanks: 0 Solve the equation... Find the x-intercepts and y-intercepts algebraically of the graph of the given equation. 18.) Solve the equation $x^{5}-3x^{3}-x^{2}-4x-1$ using your graphing calculator. Write solutions approximated to 2 decimal places. can someone explain to me in an easy to understand way on how to accomplish this. I have a TI-84 if it helps. Answer choices: A.) -1.90, -0.30, 2.10 B.) -1.86, -0.25, 2.11 C.) 0.33, 1.30 D.) 0.34, 1.30
 March 16th, 2012, 07:09 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Solve the equation... I have a TI-89 Titanium and one method for that calculator, and I assume yours will be very similar, is to enter the given function and graph it. Then go to the math menu and choose the zero option where you will be prompted to enter the lower and upper bounds, which is facilitated by allowing the user to trace along the curve and choose points on either side of the desired root, and an approximation for the root is returned. The graph makes it clear the given function has 3 real roots, and my calculator returns the values, from left to right: -1.860806, -0.2541017, 2.1149075 Another option, which is quicker because it returns all the roots at once, is to use the solve function. On my calculator, from the home screen, I go to the algebra menu and choose the solve function, and enter: solve(x^5-3x^3-x^2-4x-1=0,x) and the following is returned: x = -1.86080585311 or x = -.254101688365 or x = 2.11490754148
 March 16th, 2012, 07:42 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,104 Thanks: 1907 There should have been an "= 0" in the original post, else there's just a polynomial, not an equation. Follow-up quickie: spot the imaginary roots.
 March 16th, 2012, 09:58 PM #4 Member   Joined: Dec 2011 Posts: 44 Thanks: 0 Re: Solve the equation... ok, I got it. Thanks. For some reason, i kept looking at the instructions where it said to find the x and y intercepts. So, I kept getting a -1 in my answer, without trying to find the rest of the x-intercepts realizing this question doesn't want the y intercept. By the way, the 89 i think might be a little different than the 84. When i use the solve function on mine, it asks for a guess and gives out only 1 solution, even though there may be more than one. It'll give the closest solution to your guess.
 March 16th, 2012, 10:05 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Solve the equation... In that case, use the graph and trace function to get close to the roots for a good guess, then use those guesses and the solve function for each root.
March 17th, 2012, 04:45 AM   #6
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Re:

Quote:
 Originally Posted by skipjack Follow-up quickie: spot the imaginary roots.
$\pm i$

 March 17th, 2012, 04:52 AM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Solve the equation... Did you obtain the complex roots by factoring, or by some other method? $x^5-3x^3-x^2-4x-1=0$ $x^5+x^3-4x^3-x^2-4x-1=0$ $$$x^5+x^3$$-$$4x^3+4x$$-$$x^2+1$$=0$ $x^3$$x^2+1$$-4x$$x^2+1$$-$$x^2+1$$=0$ $$$x^2+1$$$$x^3-4x-1$$=0$ If one notices [color=#00BF00]skipjack[/color] used the term imaginary rather than complex, then if we assume a root of the form ki, where k is real we have: $(ki)^5-3(ki)^3-(ki)^2-4(ki)-1=0$ $k^5i+3k^3i+k^2-4ki-1=0$ $ki$$k^4+3k^2-4$$+$$k^2-1$$=0$ $ki$$k^2-1$$$$k^2+4$$+$$k^2-1$$=0$ $$$k^2-1$$$$ki\(k^2+4$$+1\)=0$ $k=\pm1$
March 17th, 2012, 05:22 AM   #8
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Re: Solve the equation...

Quote:
 Originally Posted by MarkFL Did you obtain the complex roots by factoring, or by some other method?
I used 'guess and check'. $i$ looked good (first guess) and $-i$ followed.

 March 17th, 2012, 06:09 PM #9 Global Moderator   Joined: Dec 2006 Posts: 20,104 Thanks: 1907 Or apply the rational root theorem.

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