1. ## Simple Algebra Problem

I am having trouble with simple algebra. It has been 27 years since I last encountered any real math, and I am in the process of reviewing to help my grandson. Can I please get step-by step instructions on how to solve this simple math problem?:

x^2+5x+9=5

Any help is appreciated.

2. Originally Posted by Emodius
I am having trouble with simple algebra. It has been 27 years since I last encountered any real math, and I am in the process of reviewing to help my grandson. Can I please get step-by step instructions on how to solve this simple math problem?:

x^2+5x+9=5

Any help is appreciated.
an equation of this type is called a "quadratic equation." there are three main ways to solve it.

1) By the quadratic formula ---will work in any case
2) By completing the square ---will work in any case
3) By foiling ---works with "nice" quadratics, but is generally simpler so use when you can

which method is your grandson currently learning?

3. Originally Posted by Emodius
I am having trouble with simple algebra. It has been 27 years since I last encountered any real math, and I am in the process of reviewing to help my grandson. Can I please get step-by step instructions on how to solve this simple math problem?:

x^2+5x+9=5

Any help is appreciated.
Three methods:
1. Factoring
x^2 + 5x + 9 = 5

x^2 + 5x + 4 = 0

Multiply the coefficient of the x^2 term (1) and the constant term (4): 1 x 4 = 4

Now write out all the factors of 4:
1, 4 <-- This also covers 4, 1
2, 2
-1, -4
-2, -2

Now look for the pair that adds to the coefficient of the linear term (5):
1 + 4 = 5
2 + 2 = 4
-1 + -4 = -5
-2 + -2 = -4

So we need the 1, 4 pair.

Now write the equation as the following:
x^2 + 5x + 4 = 0

x^2 + (1x + 4x) + 4 = 0

(x^2 + x) + (4x + 4) = 0

Now factor the common factor from each set of parenthesis:
x(x + 1) + 4(x + 1) = 0

Now each term has a common factor, the x + 1. So factor this from the overall expression:
(x + 4)(x + 1) = 0

Now, if this is to be true, one of (or both of) the factors must be 0.

Thus x + 4 = 0 ==> x = -4
or
x + 1 = 0 ==> x = -1

Thus the solution is x = -4, -1.

(This method works for any problem that factors. If it doesn't factor then you won't be able to find a pair of numbers that add to be the coefficient of the linear term. In that case you need to use one of the other two techniques.)

-Dan

4. Originally Posted by Emodius
I am having trouble with simple algebra. It has been 27 years since I last encountered any real math, and I am in the process of reviewing to help my grandson. Can I please get step-by step instructions on how to solve this simple math problem?:

x^2+5x+9=5

Any help is appreciated.
2. Completing the square.

x^2 + 5x + 9 = 5

x^2 + 5x = -4

Compare the LHS of the equation with the LHS of the following:
x^2 + 2ax + a^2 = (x + a)^2

We see that if we find the right a value then all we need to do is add a^2 to both sides to make the LHS a perfect square.

So compare the two linear coefficients:
5 = 2a ==> a = 5/2

Thus we need to add a^2 = 25/4 to both sides of the equation:
x^2 + 5x = -4

x^2 + 5x + 25/4 = -4 + 25/4

(x + 5/2)^2 = -4 + 25/4 = -16/4 + 25/4 = 9/4

(x + 5/2)^2 = 9/4

x + 5/2 = (+/-)sqrt{9/4} = (+/-)(3/2)

x = -5/2 (+/-) 3/2

So
x = -5/2 + 3/2 = -2/2 = -1
or
x = -5/2 - 3/2 = -8/2 = -4

-Dan

5. Originally Posted by Emodius
I am having trouble with simple algebra. It has been 27 years since I last encountered any real math, and I am in the process of reviewing to help my grandson. Can I please get step-by step instructions on how to solve this simple math problem?:

x^2+5x+9=5

Any help is appreciated.
x^2 + 5x + 4 = 0

If we have the trinomial ax^2 + bx + c = 0 then
x = [-b (+/-) sqrt{b^2 - 4ac}]/[2a]

In this case a = 1, b = 5, c = 4.

x = [-5 (+/-) sqrt{5^2 - 4*1*4}]/[2*1]

x = [-5 (+/-) sqrt{25 - 16}]/2

x = [-5 (+/-) sqrt{9}]/2

x = [-5 (+/-) 3]/2

So
x = [-5 + 3]/2 = [-2]/2 = -1
or
x = [-5 - 3]/2 = [-8]/2 = -4
as you might have expected by now.

-Dan

6. Use this

x= (-b +/- (b^2 -4ac)^1/2)/(2a)

7. Originally Posted by Logarithm Freak
Use this

x= (-b +/- (b^2 -4ac)^1/2)/(2a)
Don't just give an answer like this explain where it comes from, and for that
matter what the variables/symbols represent.

RonL