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## 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?

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- Thread starter b2386
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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?

- #2

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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

- #3

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Quadratic formula?????

- #4

Hootenanny

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

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- #6

Hootenanny

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

- #7

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

Thanks for the welcome^^

- #8

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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?

- #9

Hootenanny

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

When it actually simplifies to;

[tex]x = \frac{4\pm 8i}{8}[/tex]

- #10

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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..

- #11

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None of this talk about quadratics will help the OP solve their problem, as it is a cubic equation: [itex]4x^3+x+5 = 0[/itex]. 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" [Broken].

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- #12

Hootenanny

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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)

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..

- #13

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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 [Broken]

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..

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 [Broken]

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..

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- #14

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Hello..llllllllll

- #15

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@pugfug90,

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

@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) [Broken]. Doing this, you will find, as pugfug90 did, that -1 is a root.

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

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.

[tex]\text{The solution of }4x^2+4x+5=0\text{ is }x=\frac{-4\pm 8i}{8}[/tex]

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

[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

@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) [Broken]. Doing this, you will find, as pugfug90 did, that -1 is a root.

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

[tex]4x^3+x+5=(x+1)(4x^2-4x+5)=0[/tex]

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.

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- #16

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

- #17

cristo

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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.

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- #19

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

- #20

cristo

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

Because you didn't divide (x+1) into 4x

- #21

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:(

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

- #22

cristo

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Ahha!

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

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

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