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can someone please tell me how can i find the roots in these  9x^4+16x^3-76x^2-8x+32=0 numerically? since i was told it can't be done algebraically.

 Jan 18, 2016

Best Answer 

 #1
avatar+26364 
+25

can someone please tell me how can i find the roots in these  9x^4+16x^3-76x^2-8x+32=0 numerically? since i was told it can't be done algebraically.

 

http://www.wolframalpha.com/input/?i=+9x%5E4%2B16x%5E3-76x%5E2-8x%2B32%3D0

 

laugh

 Jan 18, 2016
 #1
avatar+26364 
+25
Best Answer

can someone please tell me how can i find the roots in these  9x^4+16x^3-76x^2-8x+32=0 numerically? since i was told it can't be done algebraically.

 

http://www.wolframalpha.com/input/?i=+9x%5E4%2B16x%5E3-76x%5E2-8x%2B32%3D0

 

laugh

heureka Jan 18, 2016
 #2
avatar
+10

x ~~ -3.84924

x ~~ -0.672339

x ~~ 0.658988

x ~~ 2.08481

 Jan 18, 2016
 #3
avatar+561 
+10

The easiest way to solve this is to graph it. I've used https://www.desmos.com/calculator to make the graph below.

Although you can't see the exact values of the roots on this picture, the roots are x = -3.849, -0.672, 0.659, 2.085 because these are the values of x where the line crosses the x-axis.

 Jan 18, 2016
 #4
avatar+118587 
+10

can someone please tell me how can i find the roots in these  9x^4+16x^3-76x^2-8x+32=0 numerically? since i was told it can't be done algebraically.

 

Can you do any calculus?

Just looking at it I can see it cannot have more than 4 roots.

You could just plug in some points to start with and see if you can find some negative and some positive.

If you do find one neg and on pos than there must be a zero(root) iinbetween.

You could then narrow it down - Ir would be good to use Newton's method of approximation here.

 

Really it is hard for me to tell you the best methos because I do not know wht you knowledge base is.

Drawing the graph with a graphing program like  Will and Heureka have done is a good idea.

The zeros is where it crosses the x axis. :)

 Jan 19, 2016

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