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Lesson · Grade 11 · Pre Calculus

Solving a Rational Equation and Rejecting an Extraneous Root

  • Grade 11
  • Pre Calculus
  • 2 min 13 s
  • British Columbia
  • BC11-MA-72

The video solves x²/(x − 3) = 9/(x − 3) step by step. It finds the restriction x ≠ 3 first, multiplies both sides by x − 3 to get x² = 9 and x = 3 or x = −3, rejects x = 3 as an extraneous root, warns against finding restrictions after solving or never, and checks that x = −3 gives 9/(−6) on both sides. A graph of each side then shows the curves crossing at (−3, −1.5) and both sides undefined at x = 3, and it ends on the recap "Find the restrictions first, then reject extraneous roots."

Concept: Solving a rational equation and rejecting extraneous roots

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In one sentence

The question this lesson answers

Let's solve x squared over x minus 3 equals 9 over x minus 3.

This is a rational equation: it has the variable in a denominator, the bottom of a fraction.

Transcript

What the lesson says

Let's solve x squared over x minus 3 equals 9 over x minus 3.

This is a rational equation: it has the variable in a denominator, the bottom of a fraction.

Step one: find the restrictions first, the values of x that would make a denominator zero.

x minus 3 is zero when x is 3, so 3 is a non-permissible value: x cannot equal 3.

Step two: multiply both sides by x minus 3 to clear the denominators.

The x minus 3 cancels on each side, which leaves x squared equals 9.

Take the square root of both sides to get two roots, or solutions: x equals 3 or x equals negative 3.

Step three: compare each root with the restriction. x equals 3 is not allowed, so we reject it.

It is an extraneous root: the working produces it, but it does not solve the original equation.

A common mistake is finding the restrictions after solving, or never. Then x equals 3 can slip through as an answer.

Now check x equals negative 3 in the original equation.

On the left, negative 3 squared is 9, and negative 3 minus 3 is negative 6: 9 over negative 6.

On the right, 9 over negative 3 minus 3 is also 9 over negative 6. Both sides match.

Here is a graph of each side: the left side in pink, the right side in indigo.

The curves cross at x equals negative 3, where both sides equal negative 1.5.

At x equals 3, both sides are undefined, so neither curve has a point there.

So the only solution is x equals negative 3. Find the restrictions first, then reject extraneous roots.

Curriculum

Where this fits

British Columbia, Grade 11

Pre Calculus 11, Algebra: rational expressions and equationsBC11-MA-72

Previous: Adding and Subtracting Rational ExpressionsNext: Factoring by Substitution

The outcomes and strands follow the British Columbia curriculum.

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