SOLUTION: I have these six problems that I need to work and can't remember what to do. 1. x^2-4=0 2. x^2+x=0 3. x^2+7x+12=0 4. x^2+5x=-4 5. 3x^2+5x-10=0 6. 4x^2+3x+8=0

You can put this solution on YOUR website!
You have a few options for solving these problems.

1. x^2-4=0


2. x^2+x=0


3. x^2+7x+12=0


4. x^2+5x=-4


5. 3x^2+5x-10=0


6. 4x^2+3x+8=0

---

In the first problem, you can subtract zero from both sides to get or x = +/- 2, or

you can factor the problem

---

In the second problem, you can factor x from the equation to get

, this gives us x = 0 and x = -1

---

In this problem, you have to make some observations, first notice you can factor 12 into the following pairs ({12, 1}, (6, 2} and {4, 3}). The pair {4, 3} sum to seven, so we can factor this equation as follows:

, therefore x = -4 and x = -7.

---

This problem is similar to the previous problem, but first you must subtract 4 from both sides to get x^2 + 5x + 4 = 0. The common factors of 4 are ({4,1} and {2,2}). The pair {4,1} sum to 5, so we can factor this equation as follows:

, therefore x = -1 and x = -4.

---

This problem is a bit more complex, it requires the use of the quadratic formula for finding roots of a quadratic equation with the form . If the roots are x₁ and x₂, the quadratic formula says:

x₁ = and x₂ =

For 3x^2 + 5x - 10 = 0, a = 3, b = 5 and c = -10. This means the roots are:

x₁ = , which reduces to, x₁ =
and x₂ = , which reduces to, x₂ =

---

Like the previous problem, this one requires use of the quadratic equation, unfortunately, this one involves complex numbers

The equation 4x^2 + 3x + 8 = 0 gives us a = 4, b = 3 and c = 8

The roots of this equation are:

x₁ = , which reduces to x₁ =

and x₂ = , which reduces to x₂ =

You can put this solution on YOUR website!
I have these six problems that I need to work and can't remember what to do.
1. x²-4=0

Factor the LHS as the difference of two perfect squares

(x-2)(x+2) = 0

Set each factor = 0
x-2 = 0, or x = 2
x+2 = 0, or x = -2
--------------------------------------------
2. x²+x=0

Factor the LHS by factoring out x
x(x-2)
Set each factor = 0
x=0,
x-3 = 0, or x = 3
--------------------------------------------
3. x²+7x+12=0
Factor the LHS by thinking of two positive integers that have product
12 and sum 7, they are 4 and 3.

(x+4)(x+3) = 0

Set each factor = 0
x+4 = 0, or x = 4
x+3 = 0, or x = 3
-----------------------------------
4. x²+5x=-4
Get 0 on the right by adding 4 to both sides:
x²+5x+4 = 0
Factor the LHS by thinking of two positive intergers that have product
4 and sum 5, they are 4 and 1.

(x+4)(x+3) = 0

Set each factor = 0
x+4 = 0, or x = 4
x+3 = 0, or x = 3
--------------------------------
5. 3x²+5x-10=0
That won't factor so you have to use the quadratic formula




---------------------------
6. 4x²+3x+8=0
That won't factor so you have to use the quadratic formula




That won't factor so you have to use the quadratic formula




Edwin