1
Simplifying: find the largest perfect square inside.
The squares worth knowing on sight: 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144.
72 = 36 × 2
36 is a perfect square
√72 = √36 · √2
roots split across ×
Letters work the same way — an even power is already a perfect square.
48x⁵ = 16x⁴ · 3x
16 is square, and x⁴ = (x²)²
√48x⁵ = √16x⁴ · √3x
split the product
= 4x²√3x
the leftover x stays inside
If you miss the largest square
Starting √72 with 4 × 18 gives 2√18, then 2 · 3√2 = 6√2 — the same answer, one
extra step. Half-simplified forms like 2√18 show up as choices, so keep pulling
until nothing square is left inside.
2
Roots split across × and ÷ — never across + and −.
This one rule decides half the wrong answers on this topic.
allowed
√9 · 4 = 3 · 2 = 6
√36 ÷ 4 = 6 ÷ 2 = 3
never
√9 + 16 = √25 = 5
not 3 + 4 = 7
To multiply, handle the numbers outside and the numbers inside separately — then
simplify what lands inside.
(2√6)(5√10)
outsides 2 and 5, insides 6 and 10
= 10√60
2 · 5 = 10, 6 · 10 = 60
60 = 4 · 15, so √60 = 2√15
simplify the new radicand
√72x⁵√2x = √36x⁴
dividing joins them under one root
= 6x²
the radical disappears entirely
A radical can vanish
√2 · √8 = √16 = 4. If a choice has no radical in it at all, do not rule it out —
multiplying two roots often leaves a perfect square.
3
Adding and subtracting: only like radicals combine.
Treat √2 like a letter — 3√2 + 5√2 is just 8 of them, exactly as 3y + 5y = 8y.
can combine
3√2 + 5√2 = 8√2
cannot
√2 + √3
leave it exactly as it is
√18x³ + √8x³
the opening problem — they look unlike
= 3x√2x + 2x√2x
9x² and 4x² come out
= 5x√2x
3 of them plus 2 of them
√18x³ + √8x³ = 5x√2x
Always simplify before deciding that two radicals will not combine.
The favourite distractor adds the radicands
√18x³ + √8x³ is not √26x³. Test it with the smallest numbers you like:
√9 + √16 = 3 + 4 = 7, while √25 = 5. Two different numbers, every time.
4
One root underneath: multiply top and bottom by it.
Multiplying by √6√6 is multiplying by 1 — the value does not move, only the shape.
Nothing changed but the look: 12√6 ≈ 4.899 and 2√6 ≈ 4.899. SAT answer choices are
written in the second shape, so finish the job even when the first shape is already
correct.
5
Two terms underneath: use the conjugate.
Flip the middle sign. (a + b)(a − b) = a² − b² squares both roots away at once.
10√7 + √2 · √7 − √2√7 − √2
the conjugate
bottom = (√7)² − (√2)² = 7 − 2 = 5
the two middle terms cancel
= 10(√7 − √2)5
keep the bracket whole
= 2(√7 − √2) = 2√7 − 2√2
10 ÷ 5 = 2
Only the middle sign flips
The conjugate of √7 + √2 is √7 − √2 — not −√7 − √2 and not √2 − √7. And the
denominator is 7 − 2 = 5, not 7 + 2 = 9; a choice built on 9 is always offered.
6
When the roots differ, switch to exponent form.
A square root and a cube root will not combine as radicals. As exponents they add in one line.
√x = x1/2
3√x = x1/3
xm/n = n√(xm)
(√x)(3√x) , x > 0
different roots — stuck as radicals
= x1/2 · x1/3
exponent form
12 + 13 = 36 + 26 = 56
multiplying adds
= x5/6 = ⁶√x⁵
back to whichever form the choices use
It works on stacked roots too: √√81 = 811/4 = 3, because (3)⁴ = 81.
Going the other way is just as useful: when a stem hands you x3/2 and the
choices are radicals, read it as √x³. The denominator of the exponent is the root,
the numerator is the power.
7
The traps worth naming.
| Tempting |
Correct |
Why |
| √a + b = √a + √b | false | √25 ≠ 3 + 4 |
|---|
| √ab = √a · √b | true | products are fine |
|---|
| √x² = x | |x| | √(−3)² = 3, not −3 |
|---|
| (√a)² = a | true | for a ≥ 0 |
|---|
| √2 · √8 = √16 | = 4 | a radical can vanish |
|---|
A radical answer is finished only when no perfect square is left inside and no root is
left underneath.
Almost every SAT radical stem says "where x > 0". That is what lets you write
√x² as x instead of |x| — but if the restriction is missing, the absolute value is
back.
✓
Check: the decimals have to agree.
You have a calculator for the whole section. An exact answer and its decimal are the same number — so type both and compare.
√72
8.485 vs 6(1.414) = 8.485
✓
the opener at x = 2
12 + 8 = 20 vs 5(2)√4 = 20
✓
12 ÷ √6
4.899 vs 2(2.449) = 4.899
✓
10 ÷ (√7 + √2)
2.463 vs 2(2.646 − 1.414) = 2.463
✓
the 9 trap
10(√7 − √2) ÷ 9 = 1.369
not 2.463 — ruled out ✓
Make this the habit
Store the original expression in Desmos, then test the surviving choices against it.
When a "which expression is equivalent" item carries a variable, substitute one
friendly value — x = 2 or x = 4 — into the stem and into each choice. Only the right
one matches.
Remember the flow
Squares out · Match to add · Rationalise · Confirm the decimal
One root underneath — multiply by it. Two terms — conjugate.
