A level Maths help! Solving a quartic/quadratic equation! eg. x^4 - 5x^2 + 4 = 0?

Question

I've got an A level exam in a month and I need help.

The curriculum sheet for my course says I need to know the following:

"recognise and solve equations in x which are quadratic in some form of x, eg. x^4 - 5x^2 + 4 = 0."

I'm working from memory as I can't find a book that covers the subject. I seem to remember having to substitute in: " z = x^2 ".

I can do that and then solve for z (as with a normal quadratic) and I get : z = 4, z = 1.

It's here where i get stuck.

Do I now let z = 4, therefore x = (plus or minus)sqrt 4 and
let z = 1, therefore x = (plus or minus)sqrt 1

This gives me 4 answers, which must be wrong!

Help!

Answer 1

Nope, nothing wrong. The solutions are x=-2, x=-1, x=1, and x-2.

You can factor the equation into (x^2-4)*(x^2-1).
This equals zero when x^2 = 4 and when x^2 = 1.
That gives you four solutions.

Remember, the number of roots of a polynomial can be any number up to the degree of the polynomial (i.e. the highest exponent). A 4th degree polynomial can have four roots.

Answer 2

Given the quartic polynomial equation
x^4 - 5x² + 4 = 0

Since the equation is quartic (degree 4), then naturally it has 4 solutions. This equation is a special case of the more general quartic form
ax^4 + bx³ + cx² + dx + e = 0
where b = 0 and d = 0.

This is special because it can be solved easily using quadratic methods. One is by substituting another variable (say z).
Let z = x²
Thus, z² = x^4.
x^4 - 5x² + 4 = 0

Substitute
z² - 5z + 4 = 0

Factor,
(z - 4)(z - 1) = 0

hence,
z = 4 or z = 1.

We substitute back z = x²
x² = 4 or x² = 1

Thus, the "4" solutions are
x = 2,x = -2,x = 1 or x = -1.

--------------------------------
After all, the equation can be factored as
x^4 - 5x² + 4 = 0
(x² - 4)(x² - 1) = 0
(x - 2)(x + 2)(x - 1)(x + 1) = 0

So you get the 4 solutions. I hope you get it!!!!!
^_^

Answer 3

U=X^2
U^2-5U+4=0
(U-4)(U-1)=0
X^2=4, X=+/-2, X=+/-1

Answer 4

The highest power of x is called the degree of the equation.
Number of solutions of an equation is always equal to the degree of the equation.
So the highest power of x in x^4 - 5x^2 + 4 = 0 is 4 so Degree = 4.
So number of values for x is 4.
So you should be getting 4 solutions. All 4 are right.

Answer 5

No, both your method and your solution are perfectly accurate, albeit the conclusions you derived from it are erroneous.
In the given equation, substituting z = x^2, we obtain:

z^2 - 5z+ 4 = 0
Or, (z-1)(z-4) = 0
Or, z = 1, 4
Or, x^2 = 1,4

This gives us two separate equations:
x^2-1 = 0 and x^2 - 4 = 0

Now, x^2 -1 = 0 gives:
(x+1)(x-1) = 0
Or, x = -1, 1

And, x^2 - 4 = 0 gives:
(x+2)(x-2) = 0
Or, x = -2, 2

Thus x = -1,1,-2,2.

Therefore, the problem under reference indeed admits of four solutions.

To obviate your remaining doubts on the subject you can refer to the article on "Biquadratic equations" in http://en.wikipedia.org/wiki/Quartic_equation#Biquadratic_equations

Answer 6

in short way:
x^4 - 5x^2 + 4 = 0
(x^2 -1) (x^2 -4) = 0
(x - 1) (x + 1) (x - 2) (x +2) = 0
so, x=1, x=-1, x=2 and x=-2

if you are ordered to substitute x^2 with z then z=x^2 then the equation will be
x^4 - 5x^2 + 4 = 0
z^2 - 5z + 4 =0
(z - 4)(z - 1) = 0
substitute again z with x, then
(x^2 - 4) (x^2 -1) = 0
(x - 1) (x + 1) (x - 2) (x +2) = 0
so, x=1, x=-1, x=2 and x=-2

Answer 7

This is not wrong. Your steps are right so I do not need to explain.
you can put all 4 values by the way sqrt(1) = 1 and sqrt(4) =2 and check.
Because this is an eqution of order 4 it must have 4 solutions no less and no more.
So you are absolutely Right

Answer 8

because it is a quadratic, four answers isnt a bad thing! clues in the question lol

plot the points on a graph, and youll see that there are four answer for where it crosses the x axis.....because the original equation is equal to 0, it is the same as saying it is equal to y and then subbin in 0 for y if that makes sense...basically the original equation is an equation to find out the points at which a curve crosses the x-axis

Answer 9

Substitute y=x^2 and get a quadratic like
y^2 -5y +4 = 0
then solve for y using the formula/factorisation/completing the square(which is actually the formula)
i.e for ax^2 +bx +c=0
x = {-b +or - V(b^2 -4ac)}/2a

Now the answers you get are actually those for x^2,
note that for real x, x^2>=0
So any negative answers for x^2 can be dropped, for positive answers get the square root and note that positive and negative are both possible for the roots.
Hope this clears things up!

Answer 10

no there is 4 answers plus or minus 1 and plus or minus 2, because there is 4 sloutions, the equation has degree 4 and it must be 4 solustions, its not quadratic equation.