How do you solve #8y^2-9y=-1#?
So there are several ways to go about this:
Method 1: Factoring
The first thing I'm going to do is rewrite this so it equals 0. That will be done by adding 1 to both sides.
Factoring will make this turn into a product of two linear terms, which will make this easier to solve. We will be writing this in the form of:
To figure out what each value is, the derivation comes from expanding that into:
So we want the following statements to be true:
Both values for b and d have to be either +1, or -1 in order for the statement to be true. Looking at the -9 in the center tells me they should be -1. Therefore, we can say:
So now, we have the following function:
Using the 0 product principle, which says any number multiplied by 0 is 0, we can solve for y. We just need either of them to be 0. In this case, we come with the solutions:
Method 2: Quadratic Formula
If factoring doesn't work, you can use the Quadratic Formula, which says that in a function of:
The solutions will be at:
So we'll start with our function, with 1 added to both sides:
Let's plug in the values.
And there is your answer in two ways. You could graph it if you have a calculator, but there are two methods if you don't have one.
#"express in "color(blue)"standard form";ax^2+bx+c=0#
#"using the a-c method to factor the quadratic"#
#"the factors of the product "8xx1=8#
#"which sum to - 9 are - 8 and - 1"#
#"split the middle term using these factors"#
#8y^2-8y-y+1=0larrcolor(blue)"factor by grouping"#
#"take out the "color(blue)"common factor "(y-1)#
#"equate each factor to zero and solve for "y#