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Simplify and combine rational expressions, name their excluded values, and solve rational and radical equations while checking for extraneous solutions.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson a fraction is allowed to have a polynomial on the bottom. You factor such expressions to put them in lowest terms, add and subtract them over a common denominator, and say which values of $x$ they forbid. Then you solve equations built from them, and from square roots, by clearing the fraction or the root away — and you learn why the answer that comes out is only a candidate until you have put it back into the equation you started with.
You already simplify $\frac{12}{18}$ to $\frac{2}{3}$ by cancelling the factor $6$, and you already add $\frac{1}{4} + \frac{1}{6}$ by rewriting both over $12$. You already factor $x^2 - 9$ as $(x - 3)(x + 3)$. A rational expression is a fraction whose top and bottom are polynomials, and every rule you are about to use is one of those three habits with letters in place of numbers.
Rational expression: a quotient of two polynomials, like $\frac{x^2 - 4}{x + 3}$.
Factor: something being multiplied. In $(x - 2)(x + 5)$ the factors are $x - 2$ and $x + 5$; in $x - 2$, the $x$ and the $2$ are terms, not factors.
Excluded value: a number the variable may not take, because it makes a denominator zero.
Lowest terms: no factor left in common between numerator and denominator.
Common denominator: one denominator both fractions can be written over — for expressions, the product of the denominators, or less if they share a factor.
A rational expression is one polynomial over another. It behaves exactly like a numerical fraction, with one extra duty: you must say where it is undefined.
Excluded values. Factor the denominator and set each factor to zero. $\frac{x + 1}{x^2 - 25}$ has denominator $(x - 5)(x + 5)$, so $x \ne 5$ and $x \ne -5$. Do this from the original denominator, before any cancelling.
Simplifying. Factor top and bottom completely, then cancel the factors they share: $$\frac{x^2 - 9}{x^2 + 3x} = \frac{(x - 3)(x + 3)}{x(x + 3)} = \frac{x - 3}{x}, \qquad x \ne 0,\ x \ne -3.$$ The cancelled bracket had to be non-zero for the division to be legal, which is exactly what the excluded values record.
Adding and subtracting. Find a common denominator, rewrite each fraction over it, then add the numerators only: $$\frac{1}{x} + \frac{1}{x + 2} = \frac{x + 2}{x(x + 2)} + \frac{x}{x(x + 2)} = \frac{2x + 2}{x(x + 2)}.$$ Multiplying needs no common denominator: multiply across and cancel. Dividing means multiplying by the reciprocal.
Another way: picture
Two fraction bars side by side. On the first, the bracket $(x + 3)$ appears above and below and is struck through once, leaving $\frac{x - 3}{x}$. On the second, $x + 6$ over $x$ has a stroke through each $x$ and a red cross: the $x$ on top is added, so it is no factor and the stroke is illegal.
Another way: story
A rational expression is a fraction wearing a disguise. Take the disguise off by factoring — then every rule you learned for $\frac{12}{18}$ applies unchanged, and the only new thing is remembering which values the disguise was hiding.
Cancelling a term instead of a factor. In $\frac{x + 6}{x}$ the $x$ on top is added to $6$, so nothing cancels; the expression is not $6$. Try $x = 2$: it is $4$, not $6$. Cancelling is dividing, and you may only divide away something that multiplies everything above and everything below.
Adding denominators. $\frac{1}{x} + \frac{1}{3}$ is not $\frac{2}{x + 3}$. Test $x = 3$: the left side is $\frac{2}{3}$, the right side is $\frac{1}{3}$.
Reading the excluded values off the simplified form. $\frac{x^2 - 9}{x^2 - 3x}$ simplifies to $\frac{x + 3}{x}$, but $x = 3$ is still excluded — it was banned by the expression you started with, and simplifying does not un-ban it.
Denominator $x^2 - 4x = x(x - 4)$, so $x \ne 0$ and $x \ne 4$.
Excluded values come from the original denominator, first.
Numerator $x^2 - 16 = (x - 4)(x + 4)$, a difference of squares.
Factor both lines completely before cancelling anything.
The factor $x - 4$ is shared and non-zero, so the expression is $\frac{x + 4}{x}$, with $x \ne 0$ and $x \ne 4$.
The denominators share nothing, so use their product $x(x - 5)$.
$\frac{x - 5}{x(x - 5)} + \frac{x}{x(x - 5)}$.
Each fraction is multiplied, top and bottom, by what it was missing.
Add the numerators: $\frac{2x - 5}{x(x - 5)}$, with $x \ne 0$ and $x \ne 5$.
Leave the denominator factored.
Factor: $\frac{(x - 7)(x + 7)}{x(x + 7)}$.
Cancel the shared $(x + 7)$: the answer is $\frac{x - 7}{x}$, with $x \ne 0$ and $x \ne -7$.
Simplify $\frac{x^2 - 1}{x^2 + x}$, given $x \ne 0$ and $x \ne -1$.
Answer:
Write $\frac{1}{x} + \frac{1}{x - 6}$ as a single fraction.
Answer:
Which values of $x$ must be excluded from $\frac{x + 2}{x^2 - 81}$?
Simplify $\frac{x^2 - 25}{x^2 - 5x}$, given $x \ne 0$ and $x \ne 5$.
Answer:
You already solve $3x + 5 = 17$ and $x^2 - x - 6 = 0$. You already know that $\sqrt{25} = 5$ and that squaring undoes a square root. And you have just learned where a rational expression is undefined. Solving these equations uses nothing else: you clear away the fractions or the root, solve what is left, and then check.
Rational equation: an equation with the variable in a denominator, like $\frac{4}{x} = \frac{5}{x + 2}$.
Radical equation: an equation with the variable under a root sign.
Cross-multiply: for $\frac{a}{b} = \frac{c}{d}$, replace it with $ad = bc$ — legal because it is just multiplying both sides by $bd$.
Isolate: get the root, or the fraction, alone on one side before you undo it.
Extraneous solution: a value that solves the equation you produced but not the equation you started with. It is not a mistake — it is a normal by-product that checking removes.
Both kinds are solved the same way: remove the awkward thing, solve, then check.
Rational equations. Multiply both sides by the common denominator, or, when it is one fraction equal to one fraction, cross-multiply. $$\frac{3}{x} = \frac{5}{x + 4} \;\Rightarrow\; 3(x + 4) = 5x \;\Rightarrow\; 3x + 12 = 5x \;\Rightarrow\; x = 6.$$ Then check that $x = 6$ is not an excluded value: it is not, so it stands.
Radical equations. Get the root alone, then square both sides. $$\sqrt{2x - 1} = 5 \;\Rightarrow\; 2x - 1 = 25 \;\Rightarrow\; x = 13.$$ Check: $\sqrt{25} = 5$. True.
Why checking is compulsory. Multiplying by an expression that could be zero, and squaring, are both steps that can turn a false statement into a true one: $-2 = 2$ is false, but squaring gives $4 = 4$. So the new equation may have roots the old one never had. Every candidate must be put back into the original equation, and any that fails is discarded as extraneous. A square root sign always means the non-negative root, which is why a negative candidate so often fails.
Another way: picture
A funnel. Every solution of the original equation drops into the wider equation at the top, so nothing is lost; but the wider equation holds extra values too. The check at the narrow end is the filter that lets only the genuine ones through.
Another way: story
Squaring is a door that only opens one way. Walking through it is safe, but you cannot assume everyone standing on the far side came in that way — so you ask each of them for a ticket.
Not checking. Squaring and multiplying up are one-way steps: they can only add solutions, never lose them. $\sqrt{x + 6} = x$ squares to $x^2 - x - 6 = 0$, giving $3$ and $-2$; but $\sqrt{4} = 2$, not $-2$, so only $3$ survives. Checking is part of the method, not an optional tidy-up.
Squaring term by term. $(\sqrt{x} + 2)^2$ is not $x + 4$. It is $x + 4\sqrt{x} + 4$ — so isolate the root before you square, or the root does not go away.
Keeping a solution that was excluded. Solving $\frac{x}{x - 3} = \frac{3}{x - 3}$ gives $x = 3$, but $x = 3$ makes both denominators zero. There is no solution at all.
Cross-multiply: $6(x + 3) = 8x$.
Same as multiplying both sides by $x(x + 3)$.
$6x + 18 = 8x$, so $18 = 2x$ and $x = 9$.
Check: $x = 9$ makes neither denominator zero, and $\frac{6}{9} = \frac{8}{12} = \frac{2}{3}$. Accepted.
The root already stands alone, so square: $x + 12 = x^2$.
$x^2 - x - 12 = 0$, which factors as $(x - 4)(x + 3) = 0$, giving $x = 4$ and $x = -3$.
Two candidates, not yet two solutions.
Check $x = 4$: $\sqrt{16} = 4$. True. Check $x = -3$: $\sqrt{9} = 3$, not $-3$. False.
A root sign never returns a negative number.
The only solution is $x = 4$.
Square both sides: $3x + 1 = 16$, so $3x = 15$ and $x = 5$.
Check: $\sqrt{3 \times 5 + 1} = \sqrt{16} = 4$. True, so $x = 5$.
Solve $\frac{9}{x} = \frac{11}{x + 12}$.
Answer:
Solve $\sqrt{3x - 32} = 2$.
Answer:
Squaring both sides of $\sqrt{x + 2} = x$ gives $x = 2$ and $x = -1$. Which of these actually solve the original equation?
Solve $\frac{5}{x} = \frac{11}{x + 12}$.
Answer:
One tap fills a tank in $12$ minutes; a second tap fills the same tank in $24$ minutes. Running together, how many minutes do they take?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Write $\frac{1}{x} + \frac{1}{x - 3}$ as a single fraction.
Answer:
Solve $\sqrt{3x + 30} = 6$.
Answer:
You can simplify and combine rational expressions and solve rational and radical equations, checking every answer. Simplify $\frac{x^2 - 25}{x^2 + 5x}$, and solve $\sqrt{x + 20} = x$ — saying which candidate is extraneous and why.
8. Your turn: simplify $\frac{x^2 - 49}{x^2 + 7x}$, step 2
19. Your turn: solve $\sqrt{3x + 1} = 4$, step 2