Quadratic equation (using product law)?

Question

i need the step how to do it also
1. (x-2)(x+3)=0
2. (2x-5)(2x-1)=0
3. x²+x-6=0

Answer 1

Since you asked for quadratic equation, you will have to multiply out, even thou the answer for the first two are already shown,

in the first two, multiply them out, to have them in the form:

aX² + bX + c = 0

Therefore, the first one will be:

X² + X - 6 = 0

the second one will be:

4X² - 12X + 5 = 0

and the third one will be the same:

X² + X - 6 = 0

NOW, to solve, you have to use the quadratic formula:

x1 = [ -b + SquareRoot(b² - 4ac) ] / 2a

x1 = [ -b - SquareRoot(b² - 4ac) ] / 2a

Therefore, you will have two answers, those are the two answers that will satisfy your equations.

Answer 2

(x-2)(x+3)=0
x-2=0
Add 2 to both sides so that x=2
x+3=0
Subtract 3 from both sides so x=-3
The key in solving these equations is isolating the variable.

x²+x-6=0
First, you need to factor this equation by finding a set of factors for -6 that when added together equal 1 (the coefficient of x is 1, even though you can't see it).
1 and 6 are factors of 6, but if you subtract 6 from 1 you get -1, and if you subtract 1 from 6 you get 5. The only other possible set of factors is 3 and 2. If you use a -3 and a +2, adding them togeht gives you -1, when you need 1, so the set of factors you're going to use is 3 and -2.
(x+3)(x-2)
x+3=0
x=-3
x-2=0
x=2

The answer right below me multiplies (x-2) by (x+3) and then uses the quadratic formula. Unless your homework requires using it, there's no need! Remember that the only time you should use the quadratic formula is if you cannot factor or use completing the square--I'm not sure if you have learned that yet. However, the quadratic formula is long and tedious and open to making small mistakes which can change your result.

Answer 3

look at (x-2)(x+3) = 0

set it up where it looks like this:
(x -2)
(x+3)

take the 3 and multiply it by (x-2), gives you 3x - 6

take the x of (x+3) and multiply if by (x-2), gives you x^2 - 2x

add the two results:
x^2 - 2x + 3x - 6

=x^2 + x - 6 = 0

assign a, b and c based on looking at x^2 + x - 6
a = 1 because x^2 = 1x^2
b = 1 because x = 1x
c = -6

using the formula set up:
2 * a * x = -b +- (square root of b^2 - (4 * a * c))

assign the matching values of a, b and c into the formula

2 * 1 * x = -1 +- (square root of 1^2 - (4 * 1 * -6)

2x = -1 +- (square root of 1 - (-24))

2x = -1 +- (square root of 25)

2x = -1 +- 5

2x = (-1+5), (-1-5)

2x = 4, -6

x = 2, -3