Explain how to solve quadratic equations?

Question

I have 6 different questions that I can't figure out how they got the answer. Can you get the answer for me and explain how you get to that answer. You only need to explain one if you want:
1. 6a(squared) - 4a + 2 = 0
2. 16x(squared) -24x + 9 = 0
3. 4x(squared) - 12x = -9
4. 2x(squared) + 5x - 3 = 0
5. x(squared) - x - 6 = 0

Answer 1

There's a formula you can prove by completing the square and using the identity
v^2 - w^2 = (v - w)(v + w)

ax^2 + bx + c = 0 // divide by a

x^2 + (b/a)x + c/a = 0

x^2 + 2(b/2a) + (b/2a)^2 - (b/2a)^2 + c/a = 0

[x + (b/2a)]^2 - (b^2/4a^2 - 4ac/4a^2) = 0

[x + (b/2a)]^2 - (b^2 - 4ac)/4a^2 = 0

Now, plug v=x+ (b/2a), w= sqrt (((b^2 - 4ac)/4a^2) into the last equation

[x + (b/2a) + sqrt(b^2 - 4ac)/2a][x +
+ (b/2a) - sqrt(b^2 - 4ac)/2a] = 0

or x1,2 = [-b +- sqrt(b^2 - 4ac)]/2a

Let's look at #3 for instance:
4x^2 - 12x = -9

4x^2 - 12x + 9 = 0

Plug a=4, b=-12, c=9 into the formula:

x1,2 = [12 +- sqrt(12^2 - 4*4*9)]/8 =
= [12 +- sqrt(144 - 144)]/8 = (12 +- 0)/8 = 1.5

Answer 2

There are 2 approaches. You can factor the equation or use the quadratic equation.
NOTE: I will use ^2 to mean squared since ^ is the standard operator to say raised to the power in computer programming (or in spreadsheets).
To factor, you are looking for the following:
(ax + b) * (cx + d) when we multiply that out we get

a*c*x^2 + bcx + adx + bd since you know the value of the number in front of x^2 you can figure out a and c. Similarly you can get b and d. That's usually enough to figure it out.

See http://www.purplemath.com/modules/solvquad.htm for a full description of factoring.

The other way is to use the quadratic equation which is based on this form ax^2 + bx +c

Answer = (-b +/- (b^2-4ac)^.5)/2a
read as opposite of b plus or minus the squareroot of b squared minus 4ac all over 2a.
Substitution for the first equation gives:

(4 +/- (16-48)^.5 )/12
since the square root is of a negative number, the answer to this is an irrational number. I suspect you copied it down slightly wrong unless you are late in HS or early in college.

2. 16x^2 -24x + 9 = (4x -3)*(4x-3)=0 (by factoring)
Answer 4x-3 = 0 so 4x = 3 and x = 3/4

Answer 3

You need to factor each equation using the FOIL method (first, outside, inside, last).
Remember that if the last number is negative, your factored equation will be in the form of:
(x + ?)(x - ?)
If the last number is positive, your factored equation will be either:
(x + ?)(x + ?) or (x - ?)(x - ?)
I'll explain the fourth one for you:
1) 2x^2 + 5x - 3 = 0
You need to factor 2 first. This means 2 * 1 (so your first figure in each parenthesis is 2x and 1x (because you have to multiply these two figures together and get 2x^2 as your answer. This is the FIRST part of FOIL.
Then you need to factor 3. This means 3 * 1(since the 3 is negative, the numbers have to be -3 and 1 or 3 and -1. This is the LAST part of FOIL.
The OUTSIDE and INSIDE portions mean that you multiply the outside figures and the inside figures and then ADD those two products together. In this case, the sum of the two products had to be 5x.
Then it's just a matter of figuring out what will get you to 5x. In this case, it's the following:
(2x - 1)(x + 3)
Test:
First: 2x*x = 2x^2
Outside/Inside: (2x * 3) + (-1 * x) = 6x - x = 5x
Last: -1 * 3 = -3

Good luck.

Answer 4

for the first one, you have to divide by 2
6a^2-4a+2=0
(3a^2-2a+1)=0
now you factorise in the bracket
(3a+1)(a-1)=0
either 3a+1=0
3a=-1
a=-1/3
or a-1=0
a=1

for the second one you have to factorise
16x^2-24x+9=0
(4x-3)(4x-3)=0
either 4x-3=0
4x=-3
x=-3/4
or 4x-3=0
4x=-3
x=-3/4

4x^2-12x= -9
4x^2-12x+9=0
(2x-3)(2x-3)=0
either 2x-3=0
2x=3
x=3/2
or 2x-3=0
2x=3
x=3/2

2x^2+5x-3=0
(2x-1)(x+3)=0
either 2x-1=0
2x=1
x=1/2
or x+3=0
x=-3

x^2-x-6=0
(x+2)(x-3)=0
either x+2=0
x=-2
or x-3=0
x=3