finding roots for x4 - 5x^2 + 6 = 0

[prioris]

i am reading a book that says for

x4 - 5x^2 + 6 = 0

the roots are sqrt(2), sqrt(-2), sqrt(3), sqrt(-3)

when i use the quadratic formula, the answers seem to differ

x4 - 5x2 + 6 = 0.
This equation has four roots, , and ;

-5x^2+4x+6=0

-b+/-sqrt(b^2-4ac) / 2a

-4 + sqrt(4^2-(4(-5)(6))) / -10

-4 + sqrt(16-(-120)) / -10

-b +/- sqrt(136) / -10

-4 + 11.661903789690600941748305755091 / -10 = -0.76619
-4 - 11.661903789690600941748305755091 / -10 = 1.56619

if i square the results, i don't get the result

[stapel_eliz]

x4 - 5x^2 + 6 = 0

...when i use the quadratic formula, the answers seem to differ

-5x^2+4x+6=0

Where did this new equation come from? How does it relate to the equation you're needing to solve?

Since the original equation is a quadratic in x², try using factoring, and then solve the resulting quadratic factors. This should lead quickly to the book's answer.

[prioris]

my first post was my attempt to solve it

am i doing anything wrong or is the book wrong

[stapel_eliz]

Since I don't know where you new equation came from, how it relates to the given equation, or what method you're using, I can't really say if you're doing something wrong: I don't know what you're doing. But since whatever method you're using leads to an invalid result, either it is in error, or you made a mistake at some point.

As explained previously, using the standard methods does indeed lead to the book's results, which are indeed correct. Please review what was outlined to you in that reply. Thank you!

[prioris]

>Since I don't know where you new equation came from, >how it relates to the given equation,

if a book says that 4+3=5 or 3+2=5 then it is not relevant which book it is

determining whether an answer is either right or wrong is a pure mathematical operation

fyi: i am reading "The Language of Mathematics" and it is on page 197

>or what method you're using,

http://gcse.wikia.com/wiki/Quadratic_formula

so given

-5x^2 + 4x + 6 = 0

with sqrt=square root of

the book says that the 4 roots are

sqrt(2), sqrt(-2), sqrt(3), sqrt(-3)

is this correct or not

is there a method you can use to arrive at the book answers

as opposed to my answers in first post

[Honeysuckle588]

As written, this looks like the quadratic:

But quadratic equations have at most two real solutions. You list four roots.

This leads me to think there is a typo in your equation and that it should be: . Since this is a quartic, it can indeed have four solutions.

Also, two of your listed roots are the square roots of negative numbers. This will result in complex-valued roots. But the proposed quartic which I typed in has four real roots. So I think you have typos in the listed roots as well.

Now if the proposed roots are (instead) , then these four values do solve my suggested quartic.

Finally, if you are trying to solve the quadratic that you proposed, I agree with your quadratic formula results. If you are solving the quartic then you will have to make certain adjustments (which I'm sure people here - including me would be happy to help with).

[Jesse]

There cannot be four roots. If you are dealing with a quadratic equation, then you can have 2 roots at most. You can use the discriminant to determine what type of roots you have. The discriminant is the "b^2-4ac" part of the quadratic equation. There are three cases: If b^2-4ac > 0, then there are two real roots, If b^2-4ac=0, then there is a repeated real root, and If b^2-4ac<0, then there are two complex roots.

I think the equation you mean is x^4-5x^2+6=0. In this case, you cannot use the quadratic formula since this is a quartic equation. Though an equation existis for finding roots for quartic equations,it is quite messy and impractical to memorize. The roots for this equtaion are -sqrt(2), sqrt(2), -sqrt(3), sqrt(3). These are all the roots because a quartic can have at most four roots. The quick way to find these roots is to use a graphing calculator. Using a TI-83 Plus, graph the equation and use the calc zero feature.

If you are not allowed to use a calculator, and have to find the zeros by hand then you can use this process:
1. Use the degree of the polynomial to determine the maximum number of zeros.
2. Use Descartes' Rule of Signs to determine the possible number of positive zeros and negative zeros.
3. (a) If the polynomial has integer coefficients, use the Rational Zeros Theorem to identify those rational numbers that potentially could be zeros.
3. (b) Use substitution, synthetic division, or long division to test each potential rational zero.
3. (c) Each time that a zero (and thus a factor) is found , repeat Step 3 on teh depressed equation.

[Honeysuckle588]

Hi Jesse,

A little easier in this case for the quartic. Define a new variable, such as u = x^2. Then u^2 = x^4. So now the equation can be written as:

u^2 - 5u + 6 = 0.

This is quadratic in u, and factorable. Solve for u. Then substitute x^2 into your solutions and solve the resulting two quadratics directly.

[Jesse]

That works great for this problem. That reminds me, the first thing you should do when attempting to find zeros (if possible) is to try and use factoring techniques (special products, factoring by grouping, etc.)

[stapel_eliz]

if a book says that 4+3=5 or 3+2=5 then it is not relevant which book it is

Um... yes, but I didn't ask what book you were using. I asked how -5x² + 4x + 6 related to x⁴ - 5x² + 6, and what solution method you were using.

>or what method you're using,

http://gcse.wikia.com/wiki/Quadratic_formula

so given -5x^2 + 4x + 6 = 0

How does solving this quadratic equation help you find the roots of the original quartic equation? How do these two different equations with different solutions relate? What method are you using that applies the Quadratic Formula to this different equation in order to find solutions to the original equation?

the book says that the 4 roots are sqrt(2), sqrt(-2), sqrt(3), sqrt(-3)

is this correct or not

The answer hasn't changed from the other replies: Yes, the book's answer is correct.

is there a method you can use to arrive at the book answers

Yes. The method I outlined in my first reply, which I pointed out again in a later reply and another poster applied for you, will lead to the book's answers.

(That's assuming, of course, that you are not familiar with radicals, so you don't know that and are two very different things. The latter is the correct form.)