SOLUTION: The number of real roots of the equation {(2x-3)/(x-1)}+1=(6x^2-x-6)/(x-1):

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The number of real roots of the equation {(2x-3)/(x-1)}+1=(6x^2-x-6)/(x-1):      Log On


   

Question 972977: The number of real roots of the equation
{(2x-3)/(x-1)}+1=(6x^2-x-6)/(x-1):
Found 2 solutions by macston, josgarithmetic:
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
%282x-3%29%2F%28x-1%29%2B1=%286x%5E2-x-6%29%2F%28x-1%29
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%282x-3%29%2F%28x-1%29%2B%28x-1%29%2F%28x-1%29=%286x%5E2-x-6%29%2F%28x-1%29 Multiply by (x-1).
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2x-3%2Bx-1=6x%5E2-x-6
3x-4=6x%5E2-x-6
3x=6x%5E2-x-2
0=6x%5E2-4x-2
0=%28x-1%29%286x%2B2%29
x-1=0 or 6x%2B2=0
x=1 or 6x=-2
x=1 or x=-1%2F3
Since x cannot equal 1 (division by 0)
ANSWER: x=-1/3

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
%282x-3%29%2F%28x-1%29%2B1=%286x%5E2-x-6%29%2F%28x-1%29

%282x-3%29%2F%28x-1%29%2B%28x-1%29%2F%28x-1%29=%286x%5E2-x-6%29%2F%28x-1%29

%286x%5E2-x-6-2x%2B3-x%2B1%29%2F%28x-1%29=0

%286x%5E2-4x-2%29%2F%28x-1%29=0

2%283x%5E2-2x-1%29%2F%28x-1%29=0, NOW try it. Look for how many roots for 3x%5E2-2x-1. They will either both be real or both be complex. Use the discriminant to find out.

What is the sign of %28-2%29%5E2-4%2A3%2A%28-1%29 ?
The equation either will have two real roots or it will have no real roots, and you can know this from the discriminant.

Recheck in the original equation, as the other tutor indicates. The equation will be undefined at x=1 because that independant input value creates division by zero.