# Find the real solutions of the equation help

• Oct 20th 2012, 03:42 PM
alex95
Find the real solutions of the equation help
Can someone please tell me what I am doing wrong in the following calculation:

x^4 + sqrt(3)*x^2 - 3 = 0
sqrt(3)*x^2 = 3 - x^4 // I square all terms to get rid of the root
3x^4 = 9 - x^8 // u = x^4
3u = 9 - u^2
-u^2 - 3u + 9

//Using the quadratic formula to solve this
[-(-3) +- sqrt((-3)^2 - 4(-1)(9))] / -2
[3 +- sqrt(9 + 36)] / -2
[3 +- sqrt(45)]/-2
(3 +- 6.71) / -2
u = (3 + 6.71) / -2 = -4.86
u = (3 - 6.71)/-2 = 1.86

The solution is supposed to be 1.03. I have tried to solve this several ways but somewhere I am going wrong. Any advice is really appreciated. Thanks in advance
• Oct 20th 2012, 04:06 PM
Prove It
Re: Find the real solutions of the equation help
Quote:

Originally Posted by alex95
Can someone please tell me what I am doing wrong in the following calculation:

x^4 + sqrt(3)*x^2 - 3 = 0
sqrt(3)*x^2 = 3 - x^4 // I square all terms to get rid of the root
3x^4 = 9 - x^8 // u = x^4
3u = 9 - u^2
-u^2 - 3u + 9

//Using the quadratic formula to solve this
[-(-3) +- sqrt((-3)^2 - 4(-1)(9))] / -2
[3 +- sqrt(9 + 36)] / -2
[3 +- sqrt(45)]/-2
(3 +- 6.71) / -2
u = (3 + 6.71) / -2 = -4.86
u = (3 - 6.71)/-2 = 1.86

The solution is supposed to be 1.03. I have tried to solve this several ways but somewhere I am going wrong. Any advice is really appreciated. Thanks in advance

Start by letting \displaystyle \displaystyle \begin{align*} X = x^2 \end{align*} so that your equation becomes

\displaystyle \displaystyle \begin{align*} X^2 + \sqrt{3}\,X - 3 &= 0 \\ X^2 + \sqrt{3}\,X &= 3 \\ X^2 + \sqrt{3}\,X + \left(\frac{\sqrt{3}}{2}\right)^2 &= 3 + \left(\frac{\sqrt{3}}{2}\right)^2 \\ \left( X + \frac{\sqrt{3}}{2} \right)^2 &= 3 + \frac{3}{4} \\ \left( X + \frac{\sqrt{3}}{2} \right)^2 &= \frac{15}{4} \\ X + \frac{\sqrt{3}}{2} &= \pm \frac{\sqrt{15}}{2} \\ X &= \frac{-\sqrt{3} \pm \sqrt{15}}{2} \\ x^2 &= \frac{-\sqrt{3} \pm \sqrt{15}}{2} \\ x &= \pm \sqrt{\frac{-\sqrt{3} \pm \sqrt{15}}{2}} \\ x &= \pm \sqrt{\frac{-\sqrt{3} + \sqrt{15}}{2}} \textrm{ since the other solution is nonreal} \\ x &= \pm \frac{\sqrt{2}\left(-\sqrt{3} + \sqrt{15} \right)}{2} \\ &= \pm \frac{\sqrt{30} - \sqrt{6}}{2} \end{align*}
• Oct 20th 2012, 04:40 PM
Soroban
Re: Find the real solutions of the equation help
Hello, alex95!

Quote:

$\displaystyle x^4 + \sqrt{3}x^2 - 3 \:=\: 0$

$\displaystyle \sqrt{3}x^2 \:=\: 3 - x^4$

I squared all terms to get rid of the root. . This is illegal!

$\displaystyle 3x^4 \:=\: 9 - x^8$ . This is wrong.

You must square "legally":

. . $\displaystyle \begin{array}{c}\left(\sqrt{3}x^2\right)^2 \:=\:(3-x^4)^2 \\ \\[-3mm] 9x^4 \;=\;9 - 2x^4 + x^8 \\ \\[-3mm] x^8 - 11x^4 + 9 \;=\;0 \end{array}$

Now you can apply the Quadratic Formula.

. . $\displaystyle x^4 + \sqrt{3}x^2 - 3 \;=\;0$

. . $\displaystyle x^2 \;=\;\frac{\text{-}\sqrt{3} \pm\sqrt{(\sqrt{3})^2 - 4(1)(\text{-}3)}}{2(1)} \;=\;\frac{\text{-}\sqrt{3}\pm\sqrt{15}}{2}$

Discarding the negative root, we have:

. . $\displaystyle x \;=\;\pm\sqrt{\frac{\sqrt{15}-\sqrt{3}}{2}} \;=\;\pm1.034633399$

• Oct 20th 2012, 05:39 PM
Prove It
Re: Find the real solutions of the equation help
Quote:

Originally Posted by Soroban
Hello, alex95!

You must square "legally":

. . $\displaystyle \begin{array}{c}\left(\sqrt{3}x^2\right)^2 \:=\:(3-x^4)^2 \\ \\[-3mm] 9x^4 \;=\;9 - 2x^4 + x^8 \\ \\[-3mm] x^8 - 11x^4 + 9 \;=\;0 \end{array}$

Now you can apply the Quadratic Formula.

. . $\displaystyle x^4 + \sqrt{3}x^2 - 3 \;=\;0$
. . $\displaystyle x^2 \;=\;\frac{\text{-}\sqrt{3} \pm\sqrt{(\sqrt{3})^2 - 4(1)(\text{-}3)}}{2(1)} \;=\;\frac{\text{-}\sqrt{3}\pm\sqrt{15}}{2}$
. . $\displaystyle x \;=\;\pm\sqrt{\frac{\sqrt{15}-\sqrt{3}}{2}} \;=\;\pm1.034633399$