SOLUTION: I am to solve the following equations (there are two). #1. x2(that is supposed to be squared)+3x=0 #2. x2(that is supposed to be squared)+7x+12=0 I tried doing the Zero

Question 88707: I am to solve the following equations (there are two).
#1. x2(that is supposed to be squared)+3x=0

#2. x2(that is supposed to be squared)+7x+12=0
I tried doing the Zero Factor Porperty, but i am coming up with the wrong answers.
THank you in advance for your help!
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

I'm assuming you can solve these using any technique you want right?

So let's use the quadratic formula to solve these quadratics

#1
Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve (note: since the polynomial does not have a constant term, the 3rd coefficient is zero. In other words, c=0. So that means the polynomial really looks like notice , , and )

Plug in a=1, b=3, and c=0

Square 3 to get 9

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root

Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

or

Lets look at the first part:

Add the terms in the numerator
Divide

So one answer is

Now lets look at the second part:

Subtract the terms in the numerator
Divide

So another answer is

So our solutions are:
or

Notice when we graph , we get:

and we can see that the roots are and . This verifies our answer

-----------------------------------------------------------------------------
#2

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=1, b=7, and c=12

Square 7 to get 49

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root

Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

or

Lets look at the first part:

Add the terms in the numerator
Divide

So one answer is

Now lets look at the second part:

Subtract the terms in the numerator
Divide

So another answer is

So our solutions are:
or

Notice when we graph , we get:

and we can see that the roots are and . This verifies our answer
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