Quad. equation clarification please?

Question

Quad. Equat. Claification please?
EXAMPLE:
1) x² +3x+1 = 0
2) [Sol] x²+3x=-1
3) (x+ 3/2)² -9/4=-1<--(The left side is changed to include the square of a ploynomial)
4) (x+3/2)²=4/5
5) x+3/2= √5/2
6) x=-3√5/2
*Writing a Quad. form equation in the form of (x-d)²=e is called completing a square*

My question is this, in step 3) where did the "3/2" and the "-9/4" come from?

We are started with the equation in step 1) x² -+3x+1 = 0 but we can't factor it, when and how do you pull out the two "fractions"?

Answer 1

you get 3/2 because your "b" value is 3 and you are supposed to take half of that, 3/2

the 9/4 comes from squaring the 3/2

Answer 2

The gen form of a quadratic is ax^2 + bx + c = 0

Your problem is how to factor a quadratic or solve for x using completing the square.

First, you need to check if your quadratic is factorable by solving for b^2 - 4ac

if b^2 - 4ac > 0, meaning your answer is +, it is factorable
and you have two different factors. Ex (x-2)(x+3)

if b^2 - 4ac = 0, meaning your answer is 0, it is factorable and you have two similar factors Ex (X+2)(X+2) which by the way is obtained if your quadratic is a perfect trinomial square

if b^2 - 4ac < 0, meaning your answer is -, the quadratic is said to be a non-factorable quadratic (or your answers will turn out to be complex numbers)

In your problem, X^2 + 3X + 1 = 0
a = 1, b = 3, c = 1

b^2-4ac = 3^2 - 4(1)(1) = 9-4 = 5 this is positive
so we have to find two different factors. there are several ways to do that and the most direct method is by the use of the quadratic formula.

Anyway, our task is to factor the quadratic using completing the square.

Step 1 Transpose c to the right side

x^2 + 3x = -1

Step 2 We have produced an incomplete quadratic on the left side, so we need to "complete" it as a "perfect trinomial square.

You divide b by 2 and square the result
(3/2)^2 = 9/4

Step 3 Add this to both sides of the equation

x^2 + 3x + 9/4 = -1 + 9/4

Step 4 Factor the left side and calculate the right side
The form would be (x + b/2)(x + b/2)

(x+3/2)(x+3/2) = 5/4

or (x+3/2)^2 = 5/4

Step 5 Find the square root of both sides, the square root on
the left would always be (x+b/2)

x+3/2 = sqrt (5) / sqrt (4)

Step 6 Square roots have two answers, one + and one -
Your line 4 should read = 5/4 not 4/5

Your line 6 should read
x = -3/2 + sqrt(5)/ 2 and -3/2 - sqrt(5)/2

The fastest way to solve for x or to factor is by the Quadratic Formula, any day any time :-)


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

Answer 3

I think the simplest way of looking at this is to first imagine what the equation in step 1 would look like if it could be factored as a square. It would be

x2 + 3x + 9/4 = 0

which is the same as

(x + 3/2)2

I think what is missing that would make this perfectly clear is a step between steps 2 and 3:

(x2 + 3x + 9/4) - 9/4 = -1

Answer 4

1) x² +3x+1 = 0
2) [Sol] x²+3x=-1
3)You must now add a number that completes the square to both sides of the equation. That number the coefficient of x divided by 2 and the result squared. The coefficient of x is 3 so we must add (3/2)^2 = 9/4 to both sides.
Thus x^2 +3x +9/4 = -1 +9/4
Now (x-3/2)^2 = 5/4
x-3/2 = +/- sqrt(5/4) = +/- .5(sqrt(5)
x = 1.5 +/- .5sqrt(5)

Answer 5

Find the coefficient of the middle term (3x) which is 3.
Cut that in half and you get 3/2
Square that number to get 9/4

Going from step 2 to step 3, not everything is shown
2) x^2 +3x = -1
2.1) x^2+3x (+9/4 - 9/4) = -1
they cancel each other out so the equations are equivalent
then you regroup so you can factor
2.2) (x^2+3x +9/4) - 9/4 = -1
factoring the perfect square gives you
3) (x+ 3/2)^2 -9/4 =-1

Answer 6

Step 3 comes from

x^2 + 3x + 1 = 0
x^2 + 2(3/2)x + 1 = 0
x^2 + 2(3/2)x + (3/2)^2 - (3/2)^2 + 1 = 0
(x + 3/2)^2 -(9/4) = -1