Question 38292: In class, we are learning how to use quadratic techniques to solve polynomial equations. I understand this, but in the homework there is two problems that have negative problems. I could not figure out how to do this. For example, here is one of the problems:
y^(-1)-8y^(-1/2)+12=0
I have tried many ways to do this, but can't figure it out. Thank you so much. I look forward to your replay so that I may understand how to do the other problems.
Answer by ilana(307)   (Show Source):
You can put this solution on YOUR website! Okay, this one is a bit tricky. Luckily, there's a trick!
We want it to be of the form Ax^2+Bx+C=0. So, try letting x^2=y^(-1). Then x=y^(-1/2). We are lucky in that this works perfectly- then we get x^2-8x+12=0.
Now solve for x by factoring to get x=2,6.
Now, make sure to remember we want y, not x, so if x=y^(-1/2), y=(y(-1/2))^(-2)=x^(-2), so y=1/4 and 1/36.
You can test these values in the original equation and see they work.
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