I have f(x)=(2x+1)/(x-3) - (x-1)/(x+2). I need to solve for x so that f(x)=0. How would I do this?

Question

I'm supposed to use the quadratic formula.

Okay, I got to the quadratic formula part, but the number under the radical is not a perfect square, how would I solve for x?

Answer 1

a) In your equation: put f(x)=0

b) Take the common denominator - you equation looks like:
[(2x+1)(x+2) - (x-1)(x-3)] / [(x-3)(x+2)]

c) Now use the quadratic formula as you said to resolve the numerator equation, say you find the roots, x=a & x=b

d) The final step is to make your denominator does not stay indeterminate - so "a" or "b" should not be -2 (x=-2) or 3 (x=3)

So you have your solve for x solution as the value(s) from step C that satisfy the condition (d)
I hope this helps.

Answer 2

You need to get rid of the denominators. Multiply both parts of the equation by (x-3)(x+2). This will make the equation f(x)=(2x+1)(x+2)-(x-1)(x-3). This comes out to be 2x^2+5x+2-x^2+4x-3. This simplifies to x^2+9x-1. From here, use the quadratic formula.

Answer 3

Ok, building on the equation that economicsguy came up with (I came up with the same thing) the quadratic formula will give you:
x = 81 + sqrt(85) or
x = 81 - sqrt(85)

For most math classes those would be the final exact answers.

However if you want an approximate answer (that does not include any radicals), you would enter those answers into a calculator that has a square root key, and it will give you two decimal answers that you can round to one or two significant digits.

Answer 4

(2x+1)/(x-3) - (x-1)/(x+2) = 0

[(x - 3)(x + 2)][(2x+1)/(x-3) - (x-1)/(x+2) ] = 0

(2x + 1)(x + 2) - (x - 1)(x - 3) = 0

2x² + 5x + 2 - (x² - 4x + 3) = 0

2x² + 5x + 2 - x² + 4x - 3 = 0

x² + 9x -1 = 0

x² + 9x + 81/4 - 81/4 -1 = 0

(x + 9/2)² = 85/4

x = -9/2 ± 1/2√85

Therefore:

f(x) = 0 when x = ½(-9 ± √85)

Answer 5

well anything that eaquals in 0 from multiplying one of them i zero and it dosent matter what the other one is. hope it will help.