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

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 so that your equation becomes

Re: Find the real solutions of the equation help

Hello, alex95!

You must square "legally":

. .

Now you can apply the Quadratic Formula.

I applied the Quadratic first.

. .

. .

Discarding the negative root, we have:

. .

Re: Find the real solutions of the equation help

You don't need to bother squaring both sides at all, it's ALREADY a quadratic...