Checking answers

Checking is part of solving, not an optional activity after it. A strong check asks whether an answer satisfies the original mathematics and the conditions of the question. Simply repeating the same calculation is weak because it tends to repeat the same error.

For example, if you obtain $x=4$ from an equation, substitute $4$ into the original equation. If you obtain a probability of $1.12$, reject it immediately because

$$0\leq P(A)\leq 1.$$

The best check is usually quick, independent of the main method, and capable of proving an answer wrong.

Prerequisites

You should be comfortable with exact arithmetic and rounding, calculator use, algebraic fluency and mathematical notation.

A hierarchy of checks

Use these questions in order.

1. Did I answer what was asked? Check the requested quantity, form, interval, accuracy and units.
2. Is the answer possible? Check signs, size, domain, units and contextual restrictions.
3. Does it satisfy the original problem? Substitute, differentiate, integrate or reconstruct as appropriate.
4. Can I verify it independently? Use a graph, estimate, alternative formula or reverse operation.

A check need not reproduce the entire solution. It should target the steps most likely to fail.

Check the demand before the mathematics

Many correct calculations lose marks because the final answer does not match the question.

Before moving on, underline mentally or on paper:

• the quantity required;
• whether an exact or decimal answer is wanted;
• the required significant figures or decimal places;
• any interval, such as $0\leq x<2\pi$;
• any condition, such as $x>0$ or $n\in\mathbb N$;
• required units and whether a conclusion in context is needed.

Worked example: the correct roots on a restricted interval

Solve

$$\sin x=\frac12,\qquad 0\leq x<2\pi.$$

The reference angle is $\pi/6$. Sine is positive in the first and second quadrants, so

$$x=\frac{\pi}{6},\quad \frac{5\pi}{6}.$$

Both values substitute correctly. The value $13\pi/6$ also has sine $1/2$, but it is not in the stated interval. Substitution alone is therefore not a complete check. The domain condition must also be checked.

See trigonometric equations for systematic solution methods.

Estimate before using a calculator

An estimate gives a target region and catches keying errors, misplaced brackets and unreasonable precision.

Worked example: detecting a calculator entry error

Evaluate

$$\frac{19.8^2}{0.203}.$$

Before calculating,

$$\frac{19.8^2}{0.203}\approx \frac{20^2}{0.2}=2000.$$

The calculator value is approximately

$$1931.23.$$

This is consistent with the estimate. An answer of $19.3123$ or $193123$ would signal an incorrect power of ten.

An estimate need not be very accurate. Its purpose is to distinguish a plausible scale from an impossible one.

Check equations by substitution

Substitute into the original equation, not a rearranged line near the end. Earlier rearrangement may have introduced or lost solutions.

Worked example: two quadratic roots

Suppose solving

$$2x^2-5x-3=0$$

gives $x=3$ and $x=-1/2$.

For $x=3$,

$$2(3)^2-5(3)-3=18-15-3=0.$$

For $x=-1/2$,

$$2\left(-\frac12\right)^2-5\left(-\frac12\right)-3 =\frac12+\frac52-3=0.$$

Both roots satisfy the original equation. A second check uses the sum and product of roots. For $ax^2+bx+c=0$,

$$\alpha+\beta=-\frac ba, \qquad \alpha\beta=\frac ca.$$

Here,

$$3-\frac12=\frac52=-\frac{-5}{2}, \qquad 3\left(-\frac12\right)=-\frac32=\frac{-3}{2}.$$

This checks the pair independently. Review quadratic equations if these relationships are unfamiliar.

Extraneous solutions

Some operations are not reversible without conditions. Squaring both sides can introduce solutions.

Worked example: reject a false solution

Solve

$$\sqrt{x+5}=x-1.$$

The square root is non-negative, so first $x-1\geq0$, giving $x\geq1$. Squaring gives

$$x+5=(x-1)^2=x^2-2x+1,$$

so

$$x^2-3x-4=0,$$

and therefore

$$x=4\quad\text{or}\quad x=-1.$$

Check in the original equation:

$$x=4:\quad \sqrt9=3=4-1,$$

but

$$x=-1:\quad \sqrt4=2\ne-2.$$

Thus the only solution is

$$\boxed{x=4}.$$

The value $-1$ solves the squared equation, not the original equation.

Check identities at structural and numerical levels

An identity must hold for every value in its domain. Testing one value cannot prove an identity, but it can disprove an incorrect one quickly.

Suppose you claim

$$\frac{3x+5}{(x+1)(x+2)} =\frac{2}{x+1}+\frac{1}{x+2}.$$

Recombine the right-hand side:

$$\frac{2(x+2)+(x+1)}{(x+1)(x+2)} =\frac{3x+5}{(x+1)(x+2)}.$$

This algebraic reconstruction proves the identity for $x\ne-1,-2$. A numerical spot check, say $x=0$, gives $5/2$ on both sides, but this alone would not be proof. See partial fractions for more decomposition checks.

Use inverse operations in calculus

Differentiation and integration check one another.

Worked example: check an indefinite integral

Suppose

$$\int xe^{x^2}\,dx=\frac12e^{x^2}+C.$$

Differentiate the proposed antiderivative:

$$\frac{d}{dx}\left(\frac12e^{x^2}+C\right) =\frac12e^{x^2}(2x) =xe^{x^2}.$$

The derivative reproduces the integrand, so the antiderivative is correct. The constant $C$ disappears on differentiation, which is why the check cannot determine its value.

Worked example: check a stationary point

For

$$f(x)=x^3-3x^2-9x+5,$$

suppose the stationary $x$ coordinates are $-1$ and $3$. Since

$$f'(x)=3x^2-6x-9=3(x+1)(x-3),$$

substitution gives

$$f'(-1)=0, \qquad f'(3)=0.$$

Now classify them. Since

$$f''(x)=6x-6,$$

we have $f''(-1)=-12<0$, so $x=-1$ is a local maximum, while $f''(3)=12>0$, so $x=3$ is a local minimum. Checking only that $f'(x)=0$ verifies stationarity, not the classification. Continue with stationary points and curve sketching.

Definite integrals and area

For a definite integral, check:

• sign: $\int_a^b f(x)\,dx$ is negative where negative signed area dominates;
• scale: compare with a surrounding rectangle;
• limits: reversing them changes the sign;
• area wording: geometric area cannot be negative.

For example, if $0\leq f(x)\leq4$ on $[2,5]$, then

$$0\leq\int_2^5 f(x)\,dx\leq 4(5-2)=12.$$

An answer of $17$ is impossible even before the integral is recalculated. See definite integrals and areas.

Check graphs against algebra

A sketch should agree with every algebraic feature you know:

• intercepts;
• stationary points;
• asymptotes;
• symmetry;
• domain and range;
• end behaviour.

Worked example: a rational function

Consider

$$y=\frac{2x+1}{x-3}.$$

Rewrite it as

$$y=2+\frac7{x-3}.$$

This reveals the vertical asymptote $x=3$ and horizontal asymptote $y=2$. The intercepts are

$$x=-\frac12 \quad\text{and}\quad y=-\frac13.$$

A calculator graph that appears to cross $x=3$ or settle towards $y=0$ has been entered or viewed incorrectly. A graph is evidence, not a substitute for exact algebra. Window settings and pixel resolution can conceal roots or asymptotic behaviour.

Check numerical methods by their defining property

An approximate root $x=r$ of $f(x)=0$ should make $|f(r)|$ small. However, a small residual alone can be misleading when the graph is very flat or very steep, so also check the required interval or accuracy statement.

Worked example: verify a root to three decimal places

Suppose $r=1.521$ is proposed as a root of

$$f(x)=x^3-x-2.$$

To justify a root of $1.521$ to three decimal places, test the rounding boundaries:

$$f(1.5205)\approx-0.00206,$$ $$f(1.5215)\approx0.00228.$$

Because $f$ is continuous and changes sign between these values, a root lies in

$$(1.5205,1.5215),$$

so it rounds to $1.521$. Reporting only $f(1.521)\approx-0.00069$ suggests closeness but does not by itself establish correct rounding. Learn the full argument in change of sign methods.

Check probabilities and statistics

Statistical answers have strong built-in constraints.

Probability checks

Every probability must satisfy

$$0\leq p\leq1.$$

Probabilities for mutually exclusive and exhaustive outcomes sum to $1$. Also,

$$P(A')=1-P(A),$$

and for independent events,

$$P(A\cap B)=P(A)P(B).$$

Do not use the last equality unless independence is stated or established.

Worked example: check a binomial distribution

Let $X\sim\operatorname{B}(10,0.3)$. A calculation gives

$$P(X\leq2)=0.3828.$$

The value lies in $[0,1]$, but that is only a basic check. Since

$$E(X)=np=3,$$

a probability somewhat below $1/2$ for $X\leq2$ is plausible. An independent complement check is

$$P(X\leq2)=1-P(X\geq3).$$

Both calculator routes should agree apart from rounding. If they do not, check whether the calculator endpoints are inclusive. Review the binomial distribution.

Summary statistic checks

For a data set:

• the mean must lie between the smallest and largest values;
• variance and standard deviation cannot be negative;
• adding a constant $c$ to every value adds $c$ to the mean but leaves variance unchanged;
• multiplying every value by $k$ multiplies standard deviation by $|k|$ and variance by $k^2$.

These checks often expose confusion between variance and standard deviation. See averages and spread.

Check mechanics with units and physical sense

Dimensions expose equations that cannot be physically valid. Quantities may be added or equated only when their dimensions agree.

For instance,

$$v=u+at$$

is dimensionally consistent because

$$[v]=LT^{-1}, \qquad [u]=LT^{-1}, \qquad [at]=(LT^{-2})(T)=LT^{-1}.$$

By contrast, $v=u+a$ is not dimensionally consistent.

Worked example: check a constant acceleration result

A particle starts with velocity $u=5\text{ m s}^{-1}$ and accelerates at $2\text{ m s}^{-2}$ for $4$ seconds. Suppose you obtain $v=13\text{ m s}^{-1}$ and $s=36\text{ m}$.

First,

$$v=u+at=5+2(4)=13\text{ m s}^{-1}.$$

Independently, average velocity under constant acceleration is

$$\frac{u+v}{2}=\frac{5+13}{2}=9\text{ m s}^{-1}.$$

Hence

$$s=(\text{average velocity})(\text{time})=9(4)=36\text{ m}.$$

The second method verifies the displacement without repeating the original displacement formula. See constant acceleration.

Also ask whether the sign is meaningful. A negative velocity indicates direction, not an impossible speed. A negative mass, negative elapsed time in an ordinary model, or coefficient of friction below zero is usually impossible.

Exact values, rounding and calculator displays

Keep exact values during working unless the question requires decimals. For example,

$$\frac{\sqrt2}{3}\times 6\sqrt2=4$$

exactly. Replacing $\sqrt2$ by $1.414$ too early gives only an approximation and can distort later rounding.

If a final answer is required to three significant figures:

1. retain extra calculator digits during working;
2. round once, at the end;
3. state trailing zeros when they communicate accuracy, such as $2.50$;
4. include units after rounding.

Check calculator mode as well. Trigonometric work in radians requires radian mode. For example,

$$\sin\left(\frac{\pi}{6}\right)=\frac12,$$

whereas a calculator interpreting $\pi/6$ as degrees will not return $1/2$.

Common weak checks

Repeating the same working

Reading the same lines again may miss the same sign error. Prefer substitution or a reverse operation.

Trusting the answer because it looks neat

Correct answers can be awkward, and incorrect answers can be integers. Mathematical conditions, not appearance, decide.

Treating a graph as proof

A graph can support an answer and reveal missing solutions, but it usually provides approximate evidence only.

Checking only one of several answers

Every root, coordinate, probability or case must be checked. One valid solution does not validate the others.

Ignoring assumptions

Squaring, cancelling, taking logarithms, dividing by an expression and using inverse trigonometric functions all impose conditions. Check excluded values and domains explicitly.

A fast exam routine

For a short question, use a ten-second check:

1. Read the final sentence again.
2. Check sign, scale, domain and units.
3. Substitute or reverse one decisive step.
4. Check rounding and presentation.

For a long question, use earlier parts. If part (b) gives a value for use in part (c), your result in part (b) should be compatible with it. If later working produces an impossible value, trace back to the earliest point at which a restriction was violated.

Self-check

Try these without looking back.

1. A solution of $\ln(x-2)=0$ is reported as $x=1$. Give two reasons it is wrong.
2. A student finds $\int_0^2(3x^2+1)\,dx=10$. Check the answer.
3. A probability calculation gives $P(A)=0.62$ and $P(A')=0.41$. What is wrong?
4. A particle has $u=12\text{ m s}^{-1}$, $v=4\text{ m s}^{-1}$ and constant acceleration for $3$ seconds. Check the proposed displacement $s=24\text{ m}$ independently.
5. Explain why testing $x=2$ cannot prove that two expressions are identical.

Answers

1. The domain requires $x-2>0$, so $x>2$. Also, substitution gives $\ln(-1)$, which is not defined over the reals. In fact, $\ln(x-2)=0$ gives $x-2=e^0=1$, so $x=3$.

2. An antiderivative is $x^3+x$. Therefore

   $$[x^3+x]_0^2=8+2=10,$$

   so the answer is correct. It is also plausible because the integrand is positive on $[0,2]$.

3. Complements must sum to $1$, but $0.62+0.41=1.03$. At least one value is wrong or has been rounded far too severely.

4. The average velocity is $(12+4)/2=8\text{ m s}^{-1}$. Hence $s=8(3)=24\text{ m}$, so the proposed displacement is correct.

5. Agreement at one input proves only that the expressions have the same value there. Different functions can intersect. An identity requires algebraic proof for every value in the common domain.

Next steps

Build checks into your method selection with problem solving. Then practise checking in topic-specific settings through quadratics, definite integrals, probability and mechanics modelling.