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

Question

Isaiah Anthony · Accepted Answer

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

Hazel Anglin · Answer

#y=1/8#
or
#y=1# Thus, there are various approaches to take in this: First Method: Factoring I'm going to start by rewriting this to equal 0. I'll do that by adding 1 to both sides. #8y^2-9y+1=0# This can be factored to become a product of two linear terms, which will facilitate the solution process. The result will be written as follows: #(ay-b)(cy-d)=0# The derivation comes from expanding that into: to determine each value. #(ac)y^2-(ad)y-(cb)y+bd=0# Therefore, we desire the following claims to be accurate: #ac=8# #bd=1# The statement can only be true if both b and d have values of +1 or -1. Based on the -9 in the center, I can conclude that both b and d should be -1. #(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. Thus, we currently have the following feature: #(y-1)(8y-1)=0# We can solve for y by applying the 0 product principle, which states that any number multiplied by 0 is 0. All we need is for one of them to be 0. In this instance, our solutions are as follows: #y=1# or #y=1/8# Step 2: Use the Quadratic Formula The Quadratic Formula can be used in the event that factoring is ineffective. It states that as a function of: #ay^2+by+c=0# The following address will have the solutions: #y=(-b+-sqrt(b^2-4ac))/(2a)# We'll begin by adding 1 to both sides of our function: #8y^2-9y+1=0# Now let's enter 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 you have your answer in two ways: if you have a calculator, you could graph it, but if not, there are two ways to do it.

Julia Adams · Answer

undefined To solve $8y^2 - 9y = -1$, you can use the quadratic formula or factorization.