Algebra help?

Question

Can someone help me sove x^4-6x^2+5=0 by completing the square. also, can you explain it to me?

Answer 1

If you look in your book, you'll find a fairly thorough explaination of completing the square. You can also type 'completing the square' into Google and get several milliion hits (IIRC the article in Wicki is pretty good ☺)

Next, you have to substitute some other variable (let's say z) for x² so that you get z² - 6z +5 = 0. Now you solve this (and get z = 1 and z = 5 as the roots) by completing the square and then substitute z = x² to get x² = 1 and
x² = 5 as the new roots. So you end up with x = ±√5 and x = ±1 as the answers to your original equation.

Hope that helps


Doug

Answer 2

Completing the square involves taking the middle term, and taking "half squared" of it. More explanation to come.

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

Your first goal is to take "half squared" of the coefficient of the middle term. The middle term is equal to -6. When i say "half squared", I mean multiply that value by 1/2, and then square it.
1/2 of -6 is equal to -3, -3 squared is equal to 9. So our magic number here is 9.

Now, what we're going to do is ADD the value 9 into our equation and then subtract it. The reason why we do this is because 9 - 9 = 0, and we don't want to change our equation at all.

x^4 - 6x^2 + 9 + 5 - 9 = 0

Note the placement of my 9; I put it directly after the middle term.

Now, our first three terms should be a perfect square. What we do now is note that x^4 splits into x^2 and x^2, and then we take half of the coefficient of the middle term. This is our square.

(x^2 - 3)^2 + 5 - 9 = 0
Simplify 5 - 9 now.
(x^2 - 3)^2 - 4 = 0

Bring 4 to the right hand side.
(x^2 - 3)^2 = 4

And now, take the square root of both sides. Note that when you take the square root of both sides, you have to add a "plus or minus", which I'll denote as +/-
x^2 - 3 = +/- 2
Therefore,

x^2 = 3 +/- 2, yielding two solutions

x^2 = 3 + 2
x^2 = 3 - 2

x^2 = 5
x^2 = 1

Now, we solve for each of those individually.
x = +/- sqrt(5)
x = +/- 1

So our answer for x is
x = sqrt(5), -sqrt(5), 1, -1

Answer 3

well to complete the square, they want something in the form (x - a)^4 + b = y. So when you factor that you get something like x^4 + ax^2 + a^2x + a^4 or something. And that's all I know. Ha.

But here is a simple way to look at it. You know (x^4 - 6x^2) has to equal -5 for it to equal zero. So try a few numberes, like 0 and 1 and then you see that 1 works. Then it takes some thinking but you will realize that -1 also works since all the powers are even. For you level of math, you will notice most the answers are simple whole numbers so you can always just try a few to see what you get.

Answer 4

complete the square - means that you add another constant so that the expression is (ax + b)^2

In this case, the "x" is really x^2;

Let y = x^2, and the equation becomes (y-3)^2 -4, so add 4 to both sides:


x^4 - 6x^2 + 5 + 4 = 4
(x^2 - 3) x (x^2 -3) = 4

take square root of each side

(x^2 - 3) = +- 2
x^2 = 5 OR x^ = 1

x = +- sqrt(5) OR x = +- 1

Answer 5

ok... starting from
x^4 - 6x^2 + 5 = 0

The first two terms are x^4 - 6x^2. Notice the following:
(x^2 - 3)^2 = x^4 - 6x^2 + 9
The number 9 is found by the following equation: (-6/2)^2
Now, do the following...

x^4 - 6x^2 + 9 - 9 + 5 = 0
x^4 - 6x^2 + 9 - 4 = 0
x^4 - 6x^2 + 9 = 4
(x^2 - 3)^2 = 4
x^2 - 3 = +/- 2
x^2 = 3 +/- 2 = 1 or 5
x = +/- 1 or +/- sqrt(5)

Answer 6

x^4 - 6x^2 + 5 = 0
Subtract 5 from each side
x^4 - 6x^2 = -5
Divide the coefficient of x^2 by 2
-6/2 = -3
Square it
(-3)^2 = 9
Add this number to both sides
x^4 - 6x^2 + 9 = -5 + 9
(x^2 -3)^2 = 4
Take the square root of each side
x^2 - 3 = sqrt(4) or -sqrt(4)
x^2 - 3 = 2 or x^2 - 3 = -2
For x^2 - 3 = 2, we have
x^2 = 5
x = sqrt(5) or -sqrt(5)
For x^2 - 3 = -2, we have
x^2 = 1
x = 1 or -1
Answer: -sqrt(5), sqrt(5), -1, 1