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Equivalence in practice

Solving exercises about using equivalence principles in solving equations
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Hello. Let us solve the first exercise of the equivalence in practise step. We want to solve this equation. OK, first of all, we have to compute the domain of this equation. It’s clear that here we have a problem with this denominator and also with this denominator. And we have to choose as domain the biggest subset of real numbers where these two denominators do not vanish. The domain is clearly equal to the set of real numbers minus 0 and minus 1. OK. Inside this domain, this equation always makes sense.
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Now, remember that starting with one equation, if we multiply on both sides this equation by something which is never 0 inside the domain, then we get an equation which is equivalent to the starting one. What means to be equivalent? It means that the two equations have the same set of solutions inside the domain. In such a case, it’s a good idea to multiply on both sides by x times x plus 1. What do we get? You see, if I multiply this term by x times x plus 1, I get just x plus 1. And if I multiply this term by x times x plus 1, I get x times x plus 3/2.
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And now if I multiply 0 by x times x plus 1, I get 0, OK? Let us do some just easy computations. And here we get x squared plus 1 plus 3/2 x plus 1 equals 0. That is, x squared plus 5/2 x plus 1 equals 0. And now we have a new equation which is equivalent to the starting one inside the domain. OK, let us compute the solutions of this equation of degree 2. First of all, let us compute the discriminant of this polynomial. And we get the square of 5/2, which is 25/4, minus 4 times the leading coefficient, which is 1, times the constant term, which is again 1.
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And we get, taking 4 as common denominator, 25 minus 16, which is 9/4. Therefore, the solutions of this equation are x equal to minus 5/2 plus or minus the square root of the discriminant over 2 times the leading coefficient, which is 1. That is, we get minus 5/2 plus or minus 3/2 over 2, and this is minus 5/2 plus 3/2 is minus 2/2, which is minus 1/2, which is minus 1/2. And the minus 5/2 minus 3/2 is minus 8/2, which is minus 4/2 is minus 2. OK. Now look, both these two solutions belong to the domain of our initial equation.
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Therefore, because this equation and this equation are equivalent inside this domain, we can conclude that our initial equation has minus 1/2 and minus 2 as solutions. Solutions are minus 1/2 and minus 2. Thank you for the attention.
The following exercise is solved in this step.
We invite you to try to solve it before watching the video.
In any case, you will find below a PDF file with the solution.

Exercise 1.

Solve the equation \(\dfrac 1x+\dfrac{x+\frac 32}{x+1}=0\).

Exercise 2. [Solved only in the PDF file]

Solve the equation \(\dfrac{x^4-2x^2-2x+3}{x^2-2x+1}=1\).

Exercise 3. [Solved only in the PDF file]

Solve the equation \((y+7)^2=y^2\).

Exercise 4. [Solved only in the PDF file]

Solve the equation \(\dfrac 1{x^2}+\dfrac{2x-1}{x^2-1}=0\).
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Precalculus: the Mathematics of Numbers, Functions and Equations

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