Solving for X

It's been a while, so i've forgotten how to solve this equation. I just need the information on HOW to solve it, please don't give me the answer.

Problem:
-6x^4 - 7x² - 5 = 0

Solve for X.

They're expecting you to notice that this is in quadratic form (though not, technically, a quadratic).

. . . . .-6x⁴ - 7x² - 5 = 0

Naturally, you first want to clear the leading negative by multiplying through by -1:

. . . . .6x⁴ + 7x² + 5 = 0

. . . . .6(x²)² + 7(x²) + 5 = 0

Now think about the following:

. . . . .6z² + 7z + 5 = 0

How would you solve this for "z="? Use the same method to solve for "x²=", and then solve for "x=".

Eliz.

I would just plug the numbers into the quadratic formula right?

I put it into the formula :

But i don't get any real answers. Have it done it right, or am i just way off?

Hello, iotiX!

I would just plug the numbers into the quadratic formula, right? . . . . right!

I put it into the formula: \(\displaystyle \;x\:=\:\frac{-b\,\pm\,\sqrt{b^2\,-\,4ac}}{2a}\)

But i don't get any real answers. . . . . neither do I!

Have I done it right, or am i just way off?

Click to expand...

We have the "quadratic": \(\displaystyle \;6(x^2)^2\,+\,7(x^2)\,+\,5\:=\:0\)

Then: \(\displaystyle \;x^2\;=\;\frac{-7\,\pm\,\sqrt{7^2 - 4(6)(5)}}{2(6)} \;= \;\frac{-7\,\pm\sqrt{-71}}{12}\) . . . a complex number


Where did the problem come from?
The equation you gave had all negative terms in it . . . very unusual.
. . I suspect that it came up <u>during</u> a solution.
. . Could that equation be incorrect . . . due to some earlier error?

iotiX said:

It's been a while, so i've forgotten how to solve this equation. I just need the information on HOW to solve it, please don't give me the answer.
Problem:
-6x^4 - 7x² - 5 = 0
Solve for X.

Click to expand...

After changing signs:
6x^4 + 7x^2 + 5 = 0
Since 7^2 - 4(6)(-5) = 169 (a square), I suspect a typo;
equation probably is 6x^4 + 7x^2 - 5 = 0.
That factors nicely: (2x^2 - 1)(3x^2 + 5) = 0.