Mistake Master
Rational, radical, and absolute-value equations
These three equation types are short to solve and easy to over-trust. The whole game is the step after the algebra. Squaring both sides of a radical equation can manufacture a value that solves the squared version but not the original, so $\sqrt{x+2}=x$ leads to $x=2$ and $x=-1$, yet only $x=2$ survives a check; $x=-1$ is extraneous. A rational equation hides a restriction in its denominators: $\dfrac{x}{x-3}=\dfrac{3}{x-3}+2$ clears to a value that turns out to be the one $x$ can never equal, so the equation has no solution. And an absolute-value equation splits into two cases when the right side is positive, $|2x-1|=5$ gives $x=3$ and $x=-2$, but collapses to no solution the moment the right side is negative, since an absolute value is never negative. Solve, then check the candidate, read the denominators, and open both cases. Report only what holds.
§1
What this topic is about
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Rational, radical, and absolute-value equations are quick to solve and easy to over-trust. What unites them is that the algebra alone does not finish the problem. Each one can hand you a value that looks like a solution but is not. Squaring both sides of a radical equation is not reversible, so it can introduce a value the original equation rejects. Clearing the denominators of a rational equation can hide that a value was never allowed, because it makes a denominator zero. An absolute-value equation splits into two cases, and the right side decides whether there are two solutions, one, or none. The points leak in three predictable places: reporting an extraneous solution, ignoring a domain restriction, and dropping a case in an absolute-value equation. The cure is the same each time: solve, then check the candidate against the original before you report it.
§2
Radical equations: square, then check
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A radical equation has the variable under a square root, like $\sqrt{x+2}=x$. To remove the root, you square both sides, which turns $\sqrt{x+2}=x$ into $x+2=x^{2}$. That step is legitimate, but it is not reversible: squaring sends both $5$ and $-5$ to $25$, so the squared equation can have solutions the original never did. Those extra values are extraneous. The fix is built into the method, not added on top of it: every value you find after squaring is a candidate, and you keep only the ones that satisfy the original equation.
- Isolate the radical first, then square both sides.
- Squaring is not reversible, so it can introduce values the original rejects.
- Substitute each candidate into the original. A candidate that makes the square root equal a negative number is extraneous.
Worked example. Solve $\sqrt{x+2}=x$.
Squaring gives $x+2=x^{2}$, so $x^{2}-x-2=0$, which factors as $(x-2)(x+1)=0$. The candidates are $x=2$ and $x=-1$.
$$\sqrt{2+2}=\sqrt{4}=2,$$ which matches the right side, so $x=2$ holds. But $\sqrt{-1+2}=\sqrt{1}=1$, while the right side is $-1$, and $1 \neq -1$, so $x=-1$ is extraneous. The only solution is $x=2$.
§3
Rational equations: read the denominators first
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A rational equation has the variable in a denominator, like $\dfrac{x}{x-3}=\dfrac{3}{x-3}+2$. You solve it by clearing the denominators, multiplying every term by the common denominator. But before any of that, the denominators already tell you something: the variable can never make a denominator zero, because division by zero is undefined. Here the $x-3$ in the denominator means $x$ can never equal $3$. That restriction holds no matter what the algebra later produces. If the value you solve for is the forbidden one, it is excluded, and if it was the only value, the equation has no solution.
- Before solving, set each denominator equal to zero to find the values $x$ cannot take.
- Clear the denominators and solve the resulting equation as usual.
- Compare each solution to the forbidden list. Exclude any match, even one that solved cleanly.
Worked example. Solve $\dfrac{x}{x-3}=\dfrac{3}{x-3}+2$.
The denominator $x-3$ is zero at $x=3$, so $x=3$ is not allowed. Write that down first, before any clearing.
Multiply every term by $x-3$: $x=3+2(x-3)$, so $x=3+2x-6$, which gives $-x=-3$, so $$x=3.$$ But $x=3$ is exactly the excluded value, so it cannot be a solution. This equation has no solution.
§4
Absolute-value equations: two cases, or none
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An absolute-value equation sets the distance of an expression from zero equal to a number, like $|2x-1|=5$. Because the inside can be positive or negative and still have the same absolute value, the equation opens into two cases: the inside equals $5$, or the inside equals $-5$. Solve both and keep both. The one thing to check first is the right side. An absolute value is never negative, so $|2x-1|=5$ has two solutions, $|2x-1|=0$ has exactly one, and $|2x-1|=-5$ has none at all, with no cases worth solving.
- $|A|=k$ with $k>0$ opens two cases: $A=k$ and $A=-k$.
- $|A|=0$ has one case: $A=0$.
- $|A|=k$ with $k<0$ has no solution, since an absolute value is never negative.
Worked example. Solve $|2x-1|=5$.
The right side $5$ is positive, so there are two cases: $2x-1=5$ and $2x-1=-5$.
From $2x-1=5$, $2x=6$, so $x=3$. From $2x-1=-5$, $2x=-4$, so $x=-2$. Both are solutions: $$x=3 \text{ and } x=-2.$$ Solving only the first case would drop $x=-2$.
§5
Solve, then check, then report
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All three types reward the same discipline. The algebra gives you candidates; a check decides which candidates are answers. After squaring a radical equation, test each value in the original and throw out any that fail. Before trusting a rational solution, read the denominators and exclude anything that makes one zero. With an absolute value, open both cases when the right side is positive and stop at no solution when it is negative. None of these checks takes long, and each one is the difference between the right answer and a wrong one that looks reasonable.
Worked example. Solve $\sqrt{3x+4}=x$.
Squaring gives $3x+4=x^{2}$, so $x^{2}-3x-4=0$, which factors as $(x-4)(x+1)=0$. The candidates are $x=4$ and $x=-1$.
$$\sqrt{3(4)+4}=\sqrt{16}=4,$$ so $x=4$ holds. But $\sqrt{3(-1)+4}=\sqrt{1}=1$, while the right side is $-1$, so $x=-1$ is extraneous. The solution is $x=4$.
Square a radical and check every candidate in the original; clear a rational equation only after noting which values its denominators forbid; open an absolute value into two cases when the right side is positive and none when it is negative. Solve, then check, then report only what holds.
§6
Three patterns that cost real points
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Three patterns recur on these equations. They are the same ones the diagnostic routes on.
A candidate is reported without being checked.
After squaring a radical equation, both candidates get listed even though one makes the square root equal a negative number, or only the value that looked cleaner is kept. The algebra up to the candidates is right; the missing step is the check that decides which ones survive.
Fix. Treat every value found after squaring as a candidate. Substitute each into the original equation and keep only the ones that hold. A candidate that makes a square root equal a negative number is extraneous.
A value that breaks a denominator is reported as a solution.
A rational equation is cleared and solved, and the value handed in is exactly the one that makes a denominator zero. Clearing the denominator hid the restriction; it did not remove it.
Fix. Read each denominator before solving and note the values $x$ cannot take. After solving, exclude any solution that matches. If the only value found is excluded, the equation has no solution.
An absolute-value equation loses a case, or solves one it should not.
Only the positive case is solved, so one solution goes missing; or both cases are solved even though the right side is negative, when there is really no solution at all. The bars are treated like ordinary parentheses instead of a rule that splits into cases and constrains the right side.
Fix. When an absolute value equals a positive number, open both cases and keep both values. When it equals zero there is one case. When it is set equal to a negative number, stop and report no solution, since an absolute value is never negative.
Ten quick checks across the three patterns: catching an extraneous solution, excluding a value a denominator forbids, and opening the right number of absolute-value cases. Pick or type your answer, then check. Progress is saved.