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

Algebra ->  Equations -> 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       Log On


   

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) About Me  (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

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve x%5E2%2B3%2Ax=0 (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 x%5E2%2B3%2Ax%2B0=0 notice a=1, b=3, and c=0)

x+=+%28-3+%2B-+sqrt%28+%283%29%5E2-4%2A1%2A0+%29%29%2F%282%2A1%29 Plug in a=1, b=3, and c=0



x+=+%28-3+%2B-+sqrt%28+9-4%2A1%2A0+%29%29%2F%282%2A1%29 Square 3 to get 9



x+=+%28-3+%2B-+sqrt%28+9%2B0+%29%29%2F%282%2A1%29 Multiply -4%2A0%2A1 to get 0



x+=+%28-3+%2B-+sqrt%28+9+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-3+%2B-+3%29%2F%282%2A1%29 Simplify the square root



x+=+%28-3+%2B-+3%29%2F2 Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

x+=+%28-3+%2B+3%29%2F2 or x+=+%28-3+-+3%29%2F2

Lets look at the first part:

x=0%2F2 Add the terms in the numerator
x=0 Divide

So one answer is
x=0
Now lets look at the second part:

x=-6%2F2 Subtract the terms in the numerator
x=-3 Divide

So another answer is
x=-3

So our solutions are:
x=0 or x=-3

Notice when we graph x%5E2%2B3%2Ax, we get:

+graph%28+500%2C+500%2C+-13%2C+10%2C+-13%2C+10%2C1%2Ax%5E2%2B3%2Ax%2B0%29+

and we can see that the roots are x=0 and x=-3. This verifies our answer

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#2


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve x%5E2%2B7%2Ax%2B12=0 ( notice a=1, b=7, and c=12)

x+=+%28-7+%2B-+sqrt%28+%287%29%5E2-4%2A1%2A12+%29%29%2F%282%2A1%29 Plug in a=1, b=7, and c=12



x+=+%28-7+%2B-+sqrt%28+49-4%2A1%2A12+%29%29%2F%282%2A1%29 Square 7 to get 49



x+=+%28-7+%2B-+sqrt%28+49%2B-48+%29%29%2F%282%2A1%29 Multiply -4%2A12%2A1 to get -48



x+=+%28-7+%2B-+sqrt%28+1+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-7+%2B-+1%29%2F%282%2A1%29 Simplify the square root



x+=+%28-7+%2B-+1%29%2F2 Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

x+=+%28-7+%2B+1%29%2F2 or x+=+%28-7+-+1%29%2F2

Lets look at the first part:

x=-6%2F2 Add the terms in the numerator
x=-3 Divide

So one answer is
x=-3
Now lets look at the second part:

x=-8%2F2 Subtract the terms in the numerator
x=-4 Divide

So another answer is
x=-4

So our solutions are:
x=-3 or x=-4

Notice when we graph x%5E2%2B7%2Ax%2B12, we get:

+graph%28+500%2C+500%2C+-14%2C+7%2C+-14%2C+7%2C1%2Ax%5E2%2B7%2Ax%2B12%29+

and we can see that the roots are x=-3 and x=-4. This verifies our answer