SOLUTION: Decide what values of the variable cannot possibly be solutions for the equation. Do not solve. 1x−4+1x+3=1x2−x−12 What values of x cannot be solutions of the equation?

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Question 1163693: Decide what values of the variable cannot possibly be solutions for the equation. Do not solve.
1x−4+1x+3=1x2−x−12
What values of x cannot be solutions of the equation?
Found 4 solutions by greenestamps, josgarithmetic, ikleyn, MathTherapy:
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


(1) The format of the equation is very poor; I can only guess what the equation really is. If I were to guess, it looks to me like x-4+x+3=x^2-x-12: x-4%2Bx%2B3=x%5E2-x-12

But I'm not going to spend any time on your question when I don't KNOW what the equation is.

(2) The question itself doesn't make any sense. Most equations have certain solution(s); I don't know what to make of the question "what values of x cannot be solutions" to the equation.

Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
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Do not solve.
1x−4+1x+3=1x2−x−12
What values of x cannot be solutions of the equation?
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To make better sense, write it like this: x-4%2Bx%2B3=x%5E2-x-12.
Easily simplified.
2x-1=x%5E2-x-12
0=x%5E2-3x-11
x%5E2-3x-11=0

Use your knowledge of Discriminant to help answer your question.

Answer by ikleyn(52852) About Me  (Show Source):
You can put this solution on YOUR website!
.

1x − 4 + 1x + 3 = 1x2 − x − 12


First, simplify


2x - 1 = x^2 - x - 12

x^2 - 3x - 11 = 0


Now, by applying the Rational root theorem, you can conclude that no one integer number is the solution,

and even more, no one rational number is the solution.


This conclusion works without solving equation.



Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
Decide what values of the variable cannot possibly be solutions for the equation. Do not solve.
1x−4+1x+3=1x2−x−12
What values of x cannot be solutions of the equation?


Using the discriminant, b%5E2+-+4ac, we get: 

Since 53 is > 0, and NOT a perfect square, this means, as you might know, that the ROOTS/SOLUTIONS/ZEROES of the above quadratic will be REAL, IRRATIONAL, and UNEQUAL.

While they will STILL be UNEQUAL (53+%3C%3E+0), they can NEVER be IMAGINARY or RATIONAL.