Let’s solve each problem step by step. We’re simplifying radical expressions — that means we want to take out any perfect squares from under the square root sign.
We’ll go one by one:
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1) √(125n)
Break 125 into factors: 125 = 25 × 5 → 25 is a perfect square.
So, √(125n) = √(25 × 5 × n) = √25 × √(5n) =
5√(5n)
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2) √(216v)
216 = 36 × 6 → 36 is a perfect square.
√(216v) = √(36 × 6 × v) = √36 × √(6v) =
6√(6v)
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3) √(512k²)
First, k² is already a perfect square → comes out as |k|, but since this is likely assuming positive variables, we write just
k.
Now 512: Let’s factor it. 512 ÷ 64 = 8 → 64 is a perfect square.
So 512 = 64 × 8 → √512 = √64 × √8 = 8√8
But wait — can we simplify √8? Yes! √8 = (4×2) = 2√2
So overall: √512 = 8 × 2√2 = 16√2
Then multiply by k:
16k√2
Wait — let me double-check:
Actually, better way: 512 = 256 × 2 → 256 is 16² → so √512 = √(256 × 2) = 16√2
Yes! So √(512k²) = √512 × √k² = 16√2 × k =
16k√2
✔ Correct.
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4) √(512m³)
Same as above for 512 → 16√2
Now m³ = m² × m → √m³ = √(m² × m) = m√m
So total: 16√2 × m√m =
16m√(2m)
Check: √(512m³) = √(256 × 2 × m² × m) = √256 × √m² × √(2m) = 16 × m × √(2m) =
16m√(2m) ✔
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5) √(216k⁴)
216 = 36 × 6 → √216 = 6√6
k⁴ = (k²)² → √k = k²
So: 6√6 × k² =
6k²√6
✔ Check: √(216k⁴) = √(36×6×k⁴) = √36 × √6 × √k⁴ = 6 × √6 × k² =
6k²√6
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6) √(100v³)
100 is 10² → √100 = 10
v³ = v² × v → √v³ = v√v
So: 10 × v√v =
10v√v
✔ Correct.
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7) √(80p³)
80 = 16 × 5 → √80 = 4√5
p³ = p² × p → √p³ = p√p
So: 4√5 × p√p =
4p√(5p)
✔ Check: √(80p³) = √(16×5×p²×p) = 4 × p × √(5p) =
4p√(5p)
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8) √(45p²)
45 = 9 × 5 → √45 = 3√5
p² → √p² = p
So: 3√5 × p =
3p√5
✔ Simple and correct.
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9) √(147m³)
147 = 49 × 3 → √147 = 7√3
m³ = m² × m → √m³ = m√m
So: 7√3 × m√m =
7m√(3m)
✔ Check: √(147m³) = √(49×3×m²×m) = 7 × m × √(3m) =
7m√(3m)
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10) √(100m⁴)
100 = 10² → √100 = 10
m⁴ = (m²)² → √m⁴ = m²
So: 10 × m² =
10m²
✔ No radical left — perfect!
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11) √(75x²)
75 = 25 × 3 → √75 = 5√3
x² → √x² = x
So: 5√3 × x =
5x√3
✔ Correct.
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12) √(64m⁴)
64 = 8² → √64 = 8
m⁴ → √m⁴ = m²
So: 8 × m² =
8m²
✔ Done.
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13) √(16u⁴)
16 = 4² → √16 = 4
u⁴ → √u⁴ = u²
So: 4 × u² =
4u²
✔ Easy.
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14) √(28x³)
28 = 4 × 7 → √28 = 2√7
x³ = x² × x → √x³ = x√x
So: 2√7 × x√x =
2x√(7x)
✔ Check: √(28x³) = √(4×7×x²×x) = 2 × x × √(7x) =
2x√(7x)
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All done! Now here are the final answers in order:
Final Answer:
1) 5√(5n)
2) 6√(6v)
3) 16k√2
4) 16m√(2m)
5) 6k²√6
6) 10v√v
7) 4p√(5p)
8) 3p√5
9) 7m√(3m)
10) 10m²
11) 5x√3
12) 8m²
13) 4u²
14) 2x√(7x)
Parent Tip: Review the logic above to help your child master the concept of simplify radicals worksheet answers.
