How do you solve #8y^2-9y=-1#?

2 Answers
Jun 14, 2018

Answer:

#y=1/8#
or
#y=1#

Explanation:

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.

#8y^2-9y+1=0#

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:

#(ay-b)(cy-d)=0#

To figure out what each value is, the derivation comes from expanding that into:

#(ac)y^2-(ad)y-(cb)y+bd=0#

So we want the following statements to be true:

#ac=8#
#bd=1#

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:

#(ay-1)(cy-1)=0#

Next, #ac# must be 8. They can either be 1,8 or 2,4. The answer of 1,8 is the correct one because otherwise the middle value wouldn't be 9.

So now, we have the following function:

#(y-1)(8y-1)=0#

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:

#y=1# or #y=1/8#

Method 2: Quadratic Formula

If factoring doesn't work, you can use the Quadratic Formula, which says that in a function of:

#ay^2+by+c=0#

The solutions will be at:

#y=(-b+-sqrt(b^2-4ac))/(2a)#

So we'll start with our function, with 1 added to both sides:

#8y^2-9y+1=0#

Let's plug in the values.

#y=(9+-sqrt((-9)^2-4(8)(1)))/(2(8))#

#y=(9+-sqrt(81-32))/16#

#y=(9+-sqrt(49))/16#

#y=(9+-7)/16#

#y=2/16# or #y=16/16#

#y=1/8# or #y=1#

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.

Jun 14, 2018

Answer:

#y=1/8" or "y=1#

Explanation:

#"express in "color(blue)"standard form";ax^2+bx+c=0#

#8y^2-9y+1=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"#

#color(red)(8y)(y-1)color(red)(-1)(y-1)=0#

#"take out the "color(blue)"common factor "(y-1)#

#(y-1)(color(red)(8y-1))=0#

#"equate each factor to zero and solve for "y#

#8y-1=0rArry=1/8#

#y-1=0rArry=1#