Can I solve x^4 - 4x^2 + 3 = 0 with the Quadratic Fmla?

Question

My summer hw is going over quadratic equations and this was one of them but I got stuck...
I thought I'd make
a=x^2
b=4x
c=3
but that made it worse!
Can anyone help me? You don't need to do it for me, but any hint will help! Thanks!

Answer 1

You're on the right track.

a = 1
b = -4
c = 3

The trick is, the quadratic formula in this case is

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

(Notice it's "x^2 =" instead of "x = ")

So, plugging in the numbers...

b^2 - 4ac
= (-4)^2 - 4(1)(3)
= 16 - 12
= 4

sqrt (b^2 - 4ac)
= sqrt 4
= 2

-b +- sqrt (b^2 - 4ac)
= -(-4) +- 2
= 4 +- 2
= 6 and 2

[-b +- sqrt (b^2 - 4ac)] / 2a
= (6 and 2) / 2*1
= (6 and 2) / 2
= 3 and 1

Now you have
x^2 = 3 AND x^2 = 1
take the square root of both sides...
x = +- sqrt 3 AND x = +-1

So your four answers are
-1, 1, sqrt 3, - sqrt 3
**********************************
It might be easier to see it this way...
let x^2 = w

Now you have
w^2 - 4w + 3 = 0

Solve that using quadratic formula and you get
w = 3 and w = 1

Now, substitute back the x^2 = w

x^2 = 3 and x^2 = 1
so...
x = +- sqrt 3 and x = +- 1

Answer 2

let y = x^2.
then y^2 = x^4.
so you have the new quadratic equation in y given below.
y^2 - 4y + 3 = 0
here a = 1, b = -4 and c = 3.
use the formula or factorise and get the following solution.
(y-1)(y-3) = 0 which gives y = 1 or y = 3.
if y = 1 then x^2 = 1 which gives x = -1 or +1
if y = 3 then x^2 = 3 which gives x = +sqrt(3) or - sqrt(3).
so we get all the four values of x.

Answer 3

tremendously uncomplicated. it is in simple terms algebra. take the 4x2 and situations it this is comparable to 8 and circulate it over to the magnificent and upload it to the 0.then upload the three to the 8 this is 11 then divide the nineteen by utilising 4 this is two.seventy 5 then make to make the x a great huge form make the two.seventy 5 adverse Ex. 4X2=8 -4x+3=8 3+8=11 -4x=11 -x=11/4 -x=2.seventy 5 x= -2.seventy 5 desire this helped i'm no longer the appropriate instructor/ math guy

Answer 4

Let y = x²
y² - 4y + 3 = 0
(y - 3).(y - 1) = 0
y = 1 , y = 3
x² = 1, x² = 3
x = ±1 , x = ±√3

Note
y could also have been found by using quadratic formula but not necessary to do so---easier to factorise.

Answer 5

x^(4)-4x^(2)+3=0

Since 3 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 3 from both sides.
x^(4)-4x^(2)=-3

To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.
x^(4)-4x^(2)+3=0

Since the coefficient of x^(4) is 1, the coefficients of the left-hand term in each factor must be 1.
(x^(2) )(x^(2) )=0

For a polynomial of the form x^(2)+bx+c, find two factors of c (3) that add up to b (-4). In this problem -1*-3=3 and -1-3=-4, so insert -1 as the right hand term of one factor and -3 as the right-hand term of the other factor.
(x^(2)-1)(x^(2)-3)=0

The binomial can be factored using the difference of squares formula, because both terms are perfect squares.
(x+1)(x-1)(x^(2)-3)=0

Set each of the factors of the left-hand side of the equation equal to 0.
x+1=0

Set each of the factors of the left-hand side of the equation equal to 0.
x-1=0

Set each of the factors of the left-hand side of the equation equal to 0.
x^(2)-3=0

Since 1 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 1 from both sides.
x=-1

Since -1 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 1 to both sides.
x=1

Since -3 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 3 to both sides.
x^(2)=3

Take the square root of both sides of the equation to eliminate the exponent on the left-hand side.
x=\~(3)

First, substitute in the + portion of the \ to find the first solution.
x=~(3)

Next, substitute in the - portion of the \ to find the second solution.
x=-~(3)

The complete solution is the result of both the + and - portions of the solution.
x=~(3),-~(3)

The complete solution is the set of the individual solutions.
x=-1,1,~(3),-~(3)

Answer 6

Substitute m = x^2
Then you solve
m^2 - 4m +3 = 0
factorise
(m - 1)(m - 3) = 0
solve
m = 1 or m = 3
now m = x^2
so
x^2 = 1 or x^2 = 3
therefore
x = +or-1 or x = +or- root 3

Answer 7

solve for x^2 first (a = 1, b = -4, c = 3) and then find x.

Answer 8

x^4 - 4x^2 + 3 = 0
(x^2 - 3)(x^2 - 1) = 0
solve for x:
x^2 - 3 = 0
x^2 - 1 = 0

Answer 9

x^4 - 4x^2 + 3 = 0
(x^2-1)(x^2-3)=0
(x+1)(x-1)(x^2-3)=0
x=-1 or x=1 or x^2 =3
x=-1 or x=1 or x= ±√3