SOLUTION: Struggling on this end. I am supposed to solve by introducing a substitution that transforms equation to quadratic form. (3y)^-2 + (y)^-1 -4=0 Thank you

Question 224653: Struggling on this end. I am supposed to solve by introducing a substitution that transforms equation to quadratic form.
(3y)^-2 + (y)^-1 -4=0

Thank you
Found 2 solutions by NYC Math Tutor, drj:
Answer by NYC Math Tutor(4) (Show Source): You can put this solution on YOUR website!
A quadratic equation is in the form of ax^2 + bx + c
So, what you need to do is change your exponents which are -2, -1, and 0 to 2, 1, and 0. This can be done by multiplying the exponents by -1. So, what you can do is substitute y^-1 for y. Then,
(3y^-1)^-2 + (y^-1)^-1 - 4 = 0
(3^-2)y^2 + y - 4 = 0
Answer by drj(1380) (Show Source): You can put this solution on YOUR website!
Solve by introducing a substitution that transforms equation to quadratic form.

Step 1. Let

Step 2. Substitute x into

Step 3. To solve, use the given quadratic formula below

where a=9, b=1, and c=-4

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=145 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.613421921044016, -0.724533032155128. Here's your graph:

Since , then

And , then

Step 4. The solution is 1.63 and -1.38

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit
http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit
http://www.FreedomUniversity.TV/courses/Trigonometry.

Good luck in your studies!

Respectfully,
Dr J

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