Finding the roots of an equation

[b2386]

Homework Statement

How do I find the roots of 4x^3+x+5 = 0? It doesn't appear to be in a nice form like many equations in the textbook?

 

[pugfug90]

I know one way.. graph it on your graphinc calculator. You see that -1 is a zero of the graph, therefore, (x+1)*something=4x^3+x+5..so divide 4x^3+x+5 by (x+1).. You get 4x^2 + 4x + 5..graph that.. Then you see there are no zeros, meaning that those are imaginary factors.. And I can't simplify 4x^2 + 4x + 5 but
http://www.hvks.com/Numerical/websolver.htm
says the imaginary answers are
X1=(0.5-i1)
X2=(0.5+i1)
--
PS someone remind me how to solve 4x^2 + 4x + 5 :D

 

[turdferguson]

Quadratic formula?

 

[Hootenanny]

Feldoh, although your help is appreciated please do not post complete solutions, never mind incorrect solutions, as it is contrary to our policy.

 

[Feldoh]

Sorry, and on the downside, apparently I can't add :(

 

[Hootenanny]

Sorry, and on the downside, apparently I can't add :(

No problem, don't worry bout it, your a new member so we won't send you to the gallows just yet . Seriously though, any help that you can give in the forums is very much appreciated. Welcome to the Forums!

 

[Feldoh]

No problem, don't worry bout it, your a new member so we won't send you to the gallows just yet . Seriously though, any help that you can give in the forums is very much appreciated. Welcome to the Forums!

Thanks for the welcome^^

 

[pugfug90]

Quadratic formula?

I hate the quadratic formula.

To try and simplify 4x^2 + 4x + 5... With quadratic form, I did 4(+/-)[square root of -64]/8
That simplifies to 4(+/-)/[square root of 8]i/8..

Now how do we get
X1=(0.5-i1)
X2=(0.5+i1) from that..

I prefer factoring.. by FOILING..4x^2 + 4x + 5.. Or is it impossible to deFoil imaginaries?

 

[Hootenanny]

You haven't simplified properly;

That simplifies to 4(+/-)/[square root of 8]i/8

When it actually simplifies to;

$$x = \frac{4\pm 8i}{8}$$

 

[pugfug90]

Oopd.. Forgot to take the square root after de squaring 64 :D

Is there any way to simplify 4x^2 + 4x + 5 besides quadratic formula?
I tried completing the square..
4x^2 + 4x + 5..
4(x^2 + x)=-5
(x^2 + x)=-5/4
(x^2 + x + 0.25)=-1
(x+0.5)^2=-1
x+0.5=(+/-)i
x=-0.5 (+/-)i..

I got real close.. then got that -0.5 at the end..

 

[D H]

None of this talk about quadratics will help the OP solve their problem, as it is a cubic equation: $4x^3+x+5 = 0$. The mechanics for solving cubics makes the quadratic formula look like baby stuff. Mathworld has a quite complete discussion on the cubic formula: http://mathworld.wolfram.com/CubicFormula.html" .

 

[Hootenanny]

Oopd.. Forgot to take the square root after de squaring 64 :D

Is there any way to simplify 4x^2 + 4x + 5 besides quadratic formula?
I tried completing the square..
4x^2 + 4x + 5..
4(x^2 + x)=-5
(x^2 + x)=-5/4
(x^2 + x + 0.25)=-1
(x+0.5)^2=-1
x+0.5=(+/-)i
x=-0.5 (+/-)i..

I got real close.. then got that -0.5 at the end..

This answer is actually correct, I missed the minus sign in my previous post (typo sorry ), you must have made a mistake in your previous method. As an aside completing the square is equivalent to using the quadratic equation (in fact the quadratic equation is derived by completing the square)

 

[pugfug90]

Hmm..?
http://www.hvks.com/Numerical/websolver.htm
This also yields +0.5 (+/-)i..
same as yours of
https://www.physicsforums.com/latex_images/12/1255823-0.png
But I still don't know where I went wrong with my
x=-0.5 (+/-)i.. solution from completing the square..
---
"None of this talk about quadratics will help the OP solve their problem"
Nope..:D I think the way we did it in class was to assign "p" to the lowest degree, and "q" to the highest degree, find all the factors of each, and the possible zeros are some of "p/q".. Which you can find via graphing..

 

[pugfug90]

Hello..llllllllll

 

[Tedjn]

@pugfug90,

$$\text{The solution of }4x^2+4x+5=0\text{ is }x=\frac{-4\pm 8i}{8}$$​

so your completing the square solution is absolutely correct. Remember that the quadratic formula is

$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$​

@OP,

The original equation is a cubic, and while there are clever ways to solve it using plain algebra, the easiest way without a calculator is to try every root in the Rational Root Theorem (http://www.mathwords.com/r/rational_root_theorem.htm) . Doing this, you will find, as pugfug90 did, that -1 is a root.

Once you discover that, you know that $$(x-(-1))=(x+1)$$ is a factor of $$4x^3+x+5$$ by the Factor Theorem (http://www.purplemath.com/modules/factrthm.htm) . Divide through long division or synthetic division to find that

$$4x^3+x+5=(x+1)(4x^2-4x+5)=0$$​

so the last two roots are the roots of the remaining quadratic and can be solved. In all these problems, the strategy is to try and simplify polynomials with degrees higher than two by finding easy to find roots until you get a product of quadratics which makes finding all the roots possible.

 

[pugfug90]

Does that mean that the online imaginary root calculator is wrong?

 

[cristo]

Hmm..?
http://www.hvks.com/Numerical/websolver.htm
This also yields +0.5 (+/-)i..

That website gives

For the real Polynomial:
+4x^2+4x+5
The Solutions are:
X1=(-0.5+i1)
X2=(-0.5-i1)

So no, it is not incorrect.

 

[Tedjn]

Yep, the original application of the quadratic formula a few posts up did not seem to use -b but just b.

 

[pugfug90]

How come putting the original 4x^3+x+5 doesn't decompose into -0.5..?

 

[cristo]

How come putting the original 4x^3+x+5 doesn't decompose into -0.5..?

Because you didn't divide (x+1) into 4x³+x+5 correctly. The cubic factors as 4x³+x+5=(x+1)(4x²-4x+5)

 

[pugfug90]

?
:(
So, what's right, my "completing the square", resulting in -0.5 plus blah, or the online calculator's of +0.5 plush blah

 

[cristo]

The answer given online is correct. Your completing the square method is correct, but, in post #2, you have the wrong quadratic expression. It should be 4x²-4x+5. Use your completeing the square method on this, and it should work.

 

[pugfug90]

Ahha!
So I completed the square correctly, just not the right number.. Thanks:)