Questions on Algebra: Quadratic Equation answered by real tutors!

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Question 167340: I am having trouble understanding and solving quadratic equations. x^2-2x-13=0
and 4x^2-4x+3=0
The steps I need to use are:
a. move the constant term to the right side of the equation
b. multiply each term in the equation by four times the coefficient of the x^2 term.
c. square the coefficient of the original x term and add it to both sides of the equation
d. Take the square root from both sides
e. Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.
f. set the left side of the equation equal to the negative square root of the numberon the right side of the equation and solve for x
I have to solve these kind of questions on a test and I am lost. This is what I have so far: x^2-2x-13=0
x^2 -2x=13
4x^2-8x=52
Can you help me?
~Jenn
: I am having trouble understanding and solving quadratic equations. x^2-2x-13=0
and 4x^2-4x+3=0
The steps I need to use are:
a. move the constant term to the right side of the equation
b. multiply each term in the equation by four times the coefficient of the x^2 term.
c. square the coefficient of the original x term and add it to both sides of the equation
d. Take the square root from both sides
e. Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.
f. set the left side of the equation equal to the negative square root of the numberon the right side of the equation and solve for x
I have to solve these kind of questions on a test and I am lost. This is what I have so far: x^2-2x-13=0
x^2 -2x=13
4x^2-8x=52
Can you help me?
~Jenn

Answer by nerdybill(1273) About Me  (Show Source):
You can put this solution on YOUR website!
Your method is called "completing the square"...
This site explains it quite nicely:
http://www.purplemath.com/modules/sqrquad.htm
.
x^2-2x-13=0
x^2-2x = 13
Take half the 'b' coefficient and square it:[(1/2)(-2)]^2 = [-1]^1 = 1
x^2-2x+1 = 13+1 (since you added 1 to left, do so on the right - for balance)
(x-1)^2 = 14
x-1 = sqrt(14)
x = 1(+-)sqrt(14)
.
That's 1 "plus or minus" square root of 14.
.
****************************
4x^2-4x+3=0
4x^2-4x = -3
factor the 4 on the left:
4(x^2-x) = -3
(x^2-x) = -3/4
(x^2-x+(1/4)) = -3/4 + 1/4
(x-(1/2))^2 = -2/4
x-(1/2) = sqrt(-2/4)
x = (1/2)(+-)sqrt(-1/2)
.
Since the term inside the sqrt is negative -- we have no real solutions -- rather, we have two imaginary solutions.
.
You could see the same thing using the "quadratic equation" below:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax^2+bx+c=0 (in our case 4x^2+-4x+3 = 0) has the following solutons:

x[12] = (b+-sqrt( b^2-4ac ))/2\a

For these solutions to exist, the discriminant b^2-4ac should not be a negative number.

First, we need to compute the discriminant b^2-4ac: b^2-4ac=(-4)^2-4*4*3=-32.

The discriminant -32 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -32 is + or - sqrt( 32) = 5.65685424949238.

The solution is x[12] = (--4+- i*sqrt( -32 ))/2\4 = (--4+- i*5.65685424949238)/2\4

Here's your graph:
graph( 500, 500, -10, 10, -20, 20, 4*x^2+-4*x+3 )