SOLUTION: 9r2-5r-10=0 iam trying to help my son solve this equation (9r-2)(1r-5)

Question 901893: 9r2-5r-10=0
iam trying to help my son solve this equation
(9r-2)(1r-5)
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
factor

Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .

Now multiply the first coefficient by the last term to get .

Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?

To find these two numbers, we need to list all of the factors of (the previous product).

Factors of :

1,2,3,5,6,9,10,15,18,30,45,90

-1,-2,-3,-5,-6,-9,-10,-15,-18,-30,-45,-90

Note: list the negative of each factor. This will allow us to find all possible combinations.

These factors pair up and multiply to .

1*(-90) = -90
2*(-45) = -90
3*(-30) = -90
5*(-18) = -90
6*(-15) = -90
9*(-10) = -90
(-1)*(90) = -90
(-2)*(45) = -90
(-3)*(30) = -90
(-5)*(18) = -90
(-6)*(15) = -90
(-9)*(10) = -90

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :

First Number	Second Number	Sum
1	-90	1+(-90)=-89
2	-45	2+(-45)=-43
3	-30	3+(-30)=-27
5	-18	5+(-18)=-13
6	-15	6+(-15)=-9
9	-10	9+(-10)=-1
-1	90	-1+90=89
-2	45	-2+45=43
-3	30	-3+30=27
-5	18	-5+18=13
-6	15	-6+15=9
-9	10	-9+10=1

From the table, we can see that there are no pairs of numbers which add to . So cannot be factored.

===============================================================

Answer:

So doesn't factor at all (over the rational numbers).

So is prime.

complete the square

Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics

Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert to standard form by dividing both sides by 9:
We have: . What we want to do now is to change this equation to a complete square . How can we find out values of somenumber and othernumber that would make it work?
Look at : . Since the coefficient in our equation that goes in front of r is -0.555555555555556, we know that -0.555555555555556=2*somenumber, or . So, we know that our equation can be rewritten as , and we do not yet know the other number.
We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that is equivalent to our original equation .

The highlighted red part must be equal to -1.11111111111111 (highlighted green part).

, or .
So, the equation converts to , or .

Our equation converted to a square , equated to a number (1.18827160493827).

Since the right part 1.18827160493827 is greater than zero, there are two solutions:

, or

Answer: r=1.36785649279714, -0.812300937241588.

quadratic formula

Solved by pluggable solver: Quadratic Formula

Let's use the quadratic formula to solve for r:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=9, b=-5, and c=-10

Negate -5 to get 5

Square -5 to get 25 (note: remember when you square -5, you must square the negative as well. This is because .)

Multiply to get

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

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 9 to get 18

So now the expression breaks down into two parts

or

Now break up the fraction

or

Simplify

or

So the solutions are:

or

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=385 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 1.36785649279714, -0.812300937241588. Here's your graph:

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