Tom claims √a^2=a for all real a, and hence √(-5)^2=-5. What is the error?

Worked solution for TMUA-P2-ERR2-016 is revealed on this page behind the solution details.

Why isn't the answer to TMUA-P2-ERR2-016 shown upfront?

The options are lettered without marking the correct one so you attempt TMUA-P2-ERR2-016 first — TMUA questions reward spotting the trap before you check the worked solution below.

Do I need a calculator for TMUA-P2-ERR2-016?

No — calculators are banned in the TMUA. TMUA-P2-ERR2-016 is set for a no-calculator target of about 180 seconds; the worked solution shows the fastest by-hand route.

Which TMUA specification point is TMUA-P2-ERR2-016?

TMUA-P2-ERR2-016 is Paper P2 · Common fallacies (specification code ERR2), level 2. Drill more of the same topic from this page.

TMUA-P2-ERR2-016 — Paper P2 · Common fallacies · Level 2

P2ERR2Level 2~180s target

Spec: ERR2 · No calculator · ~180s

Tom claims

\sqrt{a^2}=a

for all real

a

, and hence

\sqrt{(-5)^2}=-5

. What is the error?

Options

The computation $\sqrt{25}=5$ is wrong.
Nothing is wrong; the claim is correct.
In fact $\sqrt{a^2}=|a|$ for all real $a$ , so $\sqrt{(-5)^2}=5$ ; the sign was dropped.
$\sqrt{(-5)^2}$ is undefined over the reals.

Letters only — the answer key is hidden. Work it out, then check the solution below.

Hint (check trope)

No written hint yet — check the trope: sqrt-sign-drop, modulus-omission. Name the trap first, work the stem by hand, then open the solution below.

Show solution

The square root denotes the non-negative root, so

\sqrt{a^2}=|a|

, not

a

. For

a=-5

this is

5

. Trap: writing

\sqrt{x^2}=x

and losing the modulus.

Keep practising

Same topic · same level
P2ERR2Level 2
TMUA-P2-ERR2-004
A student solves -2x > 6 by dividing both sides by -2 to get x > -3. What is the error?
Common fallacies
Open TMUA-P2-ERR2-004 →
Same trap: sqrt-sign-drop
P2ERR2Level 1
TMUA-P2-ERR2-012
Mia solves x^2=9 and concludes x=3. What is the error?
Common fallacies
Open TMUA-P2-ERR2-012 →
Step up a level
P2ERR2Level 3
TMUA-P2-ERR2-009
A sixth-former optimises a profit function P(x) and argues as follows: Fermat's theorem says that if P has a local maxim…
Common fallacies
Open TMUA-P2-ERR2-009 →

Browse more

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Options

The computation

\sqrt{25}=5

is wrong.

Nothing is wrong; the claim is correct.

In fact

\sqrt{a^2}=|a|

for all real

a

, so

\sqrt{(-5)^2}=5

; the sign was dropped.

\sqrt{(-5)^2}

is undefined over the reals.

Letters only — the answer key is hidden. Work it out, then check the solution below.

Keep practising

Same topic · same level

P2ERR2Level 2

TMUA-P2-ERR2-004

A student solves -2x > 6 by dividing both sides by -2 to get x > -3. What is the error?

Common fallacies

Open TMUA-P2-ERR2-004 →

Same trap: sqrt-sign-drop

P2ERR2Level 1

TMUA-P2-ERR2-012

Mia solves x^2=9 and concludes x=3. What is the error?

Common fallacies

Open TMUA-P2-ERR2-012 →

Step up a level

P2ERR2Level 3

TMUA-P2-ERR2-009

A sixth-former optimises a profit function P(x) and argues as follows: Fermat's theorem says that if P has a local maxim…

Common fallacies

Open TMUA-P2-ERR2-009 →

TMUA-P2-ERR2-016 — Paper P2 · Common fallacies · Level 2

Options

Keep practising

Master Common fallacies

TMUA-P2-ERR2-016 — Paper P2 · Common fallacies · Level 2

Options

Keep practising

Master Common fallacies