Quadratic equation__help__?

to solve for X use the quadratic formula.

its to complicated to write in here ill send you a link to it and try to memorize it, it can be used for all quadratic equations

http://en.wikipedia.org/wiki/Quadratic_equation

just sub a=4, b= - 6, C=1 into the quadratic formula to get your 2 x values hope this helped

use the quadratic formula with a=4, b=-6, c=1

The quadratic formula is hard to type with a keyboard so I've provided a link to it below.

The best way would be the quadratic formula. You will notice that your quadratic comes in the form of:

a² + bx + c = 0

a = 4
b = -6
c = 1

Plug these values into the quadratic formula and you will get your two answers:

x = [-b ± √(b² - 4ac)] / 2a

x = [6 ± √((-6)² - 4*4*1)] / 2*4
x = [6 ± √(36 - 16)] / 8
x = (6 ± √20) / 8
x = (6 ± 2√5) / 8

x = (3 ± √5) / 4

If you can forgive my inability to type the classical way of expressing such terms as 'square root'. and 'plus or minus' then here goes.

The simple method, as other answers indicate, is to use the formula.
What if you cannot remember it? Then we use the method known as
'Completing the square' as follows:-

4x^2 - 6x + 1 = 0
Divide by 4 so that the coefficient of x^2 is equal to 1.
x^2 - (6/4)x + 1/4 = 0
Rearrange so that the LHS only has terms involving x.
x^2 - (3/2)x = -(1/4)

Now the vital part. Try to remember this as a sentence.

Add to EACH side the SQUARE of HALF the coeffient of x. Here it is (3/4)^2.
x^2 - (3/2)x + (3/4)^2 = -(1/4) + (3/4)^2

x^2 - (3/2)x + 9/16 = -(1/4) + 9/16 (RHS is -(4/16 + (9/16)

LHS is now a 'perfect square' so we can write

(x- 3/4)^2 = 5/16 See note below.

Take square root of both sides remembering the + or -.

x-3/4 = + or - (square root(5/16))

giving x = 3/4 + (square root 5)/4 or 3/4 - (square root 5)/4.

Points to remember:-
1, At start, make coefficient of squared term equal to unity.
2. What to add to both sides to 'complete the square'.

URGENT note.
In above, if RHS is a NEGATIVE value you will need to have studied Complex Numbers in order to continue but the method is still the same.

Hope this helps you. If so then share it with your friends if you want to.

4x^2 - 6x + 1 = 0
x = [-b ±√(b^2 - 4ac)]/2a (<== quadratic formula)

a = 4
b = -6
c = 1

x = [6 ±√(36 - 16)]/8
x = [6 ±√20]/8
x ≈ [6 ±4.42]/8

x ≈ [6 + 4.42]/8
x ≈ 10.42/8
x ≈ 1.3025

x ≈ [6 - 4.42]/8
x ≈ 1.58/8
x ≈ 0.1975

∴ x ≈ 0.1975 , 1.3025

Use the quadratic formula using a= 4, b= -6 and c= 1- following this formula should help you solve the equation.

hope dis helps....

http://www.hyper-ad.com/tutoring/math/algebra/Quadratic%20Formula%20Examples.html

If Ax^2 + Bx + C=0,then there are two x's;
X1=[-B+(B^2-4AC)^1/2] / 2A & X2=[-B-(B^2-4AC)^1/2] / 2A
in your particular case,X1=[6+(36-16)^1/2] / 8=1.309 & X2=0.19

4x² - 6x + 1 = 0 is the question.

Now, use the Quadratic Formula.

x = [-b ± √(b² - 4ac)] / 2a:
where b = -6, a = 4, c = 1.

=> x = [6 ± √(36 - 16)] / 8

=> x = [6 ± √20] / 8

=> x = [6 ± 2√5] / 8

=> x = [6 + 2√5] / 8 OR [6 - 2√5] / 8

=> x = ¾ + (√5/4) OR ¾ - (√5/4)

So, x = ¾ ± (√5/4)

Hope that helped....