Find all the real roots of the equation :

$\displaystyle

2x^6+x^4-8x^2+4=0

$

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- Jun 10th 2009, 11:44 PMthereddevilsfactor theorem
Find all the real roots of the equation :

$\displaystyle

2x^6+x^4-8x^2+4=0

$ - Jun 11th 2009, 01:05 AMAmer

the possible root is $\displaystyle \pm 2 , \pm 1,\pm 4 , \pm \frac{1}{2} $

but no one of the possible root is a root of the polynomial that you wrote

there is no real root can be found without calculator see what I have when I put them in the calculator I think there is wrong in your equation

Attachment 11857

Edited : - correct something yeongil mentioned - Jun 11th 2009, 01:17 AMyeongil
- Jun 11th 2009, 05:36 AMthereddevils
Btw the answer is $\displaystyle \pm\sqrt{2}$

- Jun 11th 2009, 06:37 AMyeongil
No, they are

**not**. If you plug either $\displaystyle \sqrt{2}$ or $\displaystyle -\sqrt{2}$ into the original equation:

$\displaystyle 2x^6+x^4-8x^2+4=0$

you get 8 = 0, so neither solution works.

However, if your original equation was supposed to be

$\displaystyle 2x^6 + x^4 - 8x^2 {\color{red}-} 4 = 0$

then yes, those two solutions work. In fact, it makes solving this equation much easier, as you can solve it by factoring by grouping:

$\displaystyle \begin{aligned}

2x^6 + x^4 - 8x^2 - 4 &= 0 \\

x^4(2x^2 + 1) - 4(2x^2 + 1) &= 0 \\

(x^4 - 4)(2x^2 + 1) &= 0 \\

(x^2 - 2)(x^2 + 2)(2x^2 + 1) &= 0

\end{aligned}$

Set each factor equal to zero:

$\displaystyle \begin{aligned}

x^2 - 2 &= 0 \\

x^2 &= 2 \\

x &= \pm \sqrt{2}

\end{aligned}$

(the two real solutions)

$\displaystyle \begin{aligned}

x^2 + 2 &= 0 \\

x^2 &= -2 \\

x &= \pm i\sqrt{2}

\end{aligned}$

(two non-real solutions)

$\displaystyle \begin{aligned}

2x^2 + 1 &= 0 \\

2x^2 &= -1 \\

x^2 &= -\frac{1}{2} \\

x &= \pm \frac{\sqrt{2}}{2}i

\end{aligned}$

(two more non-real solutions)

So it looks like you copied the problem wrong? :confused:

01 - Jun 11th 2009, 06:50 AMthereddevils
oh thanks a lot .. nope , i copied it down correctly .. i think its the book's mistake

Btw i hv a question here .. For questions like this where the leading coefficient is not 1 , is it a must for me to reduce the coefficient to 1 before i find the factors : example in this case

$\displaystyle 2x^6+x^4-8x^2+4=0$

$\displaystyle x^6+\frac{1}{2}x^4-4x^2+2=0$ , is this step necessary

Before i start substituting the factors of the constant term to find the factors of the equation .

THanks . - Jun 12th 2009, 03:44 AMgreat_math