Square root & cube root equations--solving & graphing radicals | TPT - Free Printable
Educational worksheet: Square root & cube root equations--solving & graphing radicals | TPT. Download and print for classroom or home learning activities.
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Step-by-step solution for: Square root & cube root equations--solving & graphing radicals | TPT
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Show Answer Key & Explanations
Step-by-step solution for: Square root & cube root equations--solving & graphing radicals | TPT
Let’s solve each radical equation step by step. We’ll isolate the radical, square both sides to eliminate it, and then solve for x. Always check your answer in the original equation — sometimes squaring introduces “fake” solutions that don’t actually work!
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1. √(x² + 4) = 5
Square both sides:
x² + 4 = 25
Subtract 4:
x² = 21
Take square root:
x = ±√21
Check:
Plug back into original: √((±√21)² + 4) = √(21 + 4) = √25 = 5 → ✔ Works for both.
✔ Final Answer: x = √21 or x = -√21
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2. √(8x - 1) + 9 = 0
Isolate radical:
√(8x - 1) = -9
But wait — square roots are always ≥ 0 (in real numbers). You can’t have a square root equal to -9. So no solution.
✔ Final Answer: No solution
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3. √(9y - 7) - 1 = 5
Add 1 to both sides:
√(9y - 7) = 6
Square both sides:
9y - 7 = 36
Add 7:
9y = 43
Divide by 9:
y = 43/9
Check:
√(9*(43/9) - 7) - 1 = √(43 - 7) - 1 = √36 - 1 = 6 - 1 = 5 → ✔
✔ Final Answer: y = 43/9
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4. √(3x) = √(3x - 1)
Square both sides:
3x = 3x - 1
Subtract 3x from both sides:
0 = -1 → ✘ Contradiction!
No value of x makes this true.
✔ Final Answer: No solution
---
5. √(t + 1) = √(t - 1)
Square both sides:
t + 1 = t - 1
Subtract t:
1 = -1 → ✘ Contradiction!
No solution.
✔ Final Answer: No solution
---
6. x = √(x² + 4)
Square both sides:
x² = x² + 4
Subtract x²:
0 = 4 → ✘ Contradiction!
Wait — but let’s think: if x is positive, maybe? Try plugging in values.
Try x = 2: √(4 + 4) = √8 ≈ 2.8 ≠ 2
Try x = 3: √(9+4)=√13≈3.6≠3
Actually, since √(x² + 4) > |x| for all real x, it can never equal x unless... never.
So no solution.
✔ Final Answer: No solution
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7. 4 + √(2t - 2) = 4
Subtract 4:
√(2t - 2) = 0
Square both sides:
2t - 2 = 0
2t = 2
t = 1
Check: 4 + √(2*1 - 2) = 4 + √0 = 4 → ✔
✔ Final Answer: t = 1
---
8. z + 4 = √(3z + 4)
Isolate radical? It’s already on one side. Square both sides:
(z + 4)² = 3z + 4
Expand left: z² + 8z + 16 = 3z + 4
Bring all to left: z² + 5z + 12 = 0
Discriminant: 25 - 48 = -23 → No real solutions.
Wait — did I do that right?
Left: (z+4)^2 = z² + 8z + 16
Right: 3z + 4
So: z² + 8z + 16 - 3z - 4 = 0 → z² + 5z + 12 = 0 → yes.
Discriminant negative → no real solution.
But let’s test small values just in case.
Try z=0: 0+4=4; √(0+4)=2 → 4≠2
z=1: 5 vs √7≈2.6 → no
z=-1: 3 vs √1=1 → no
z=-3: 1 vs √(-9+4)=√(-5) → invalid
So indeed, no real solution.
✔ Final Answer: No real solution
---
9. √(9 - 7k) = √(1 - k)
Square both sides:
9 - 7k = 1 - k
Add 7k to both sides:
9 = 1 + 6k
Subtract 1:
8 = 6k
k = 8/6 = 4/3
Check:
Left: √(9 - 7*(4/3)) = √(9 - 28/3) = √(27/3 - 28/3) = √(-1/3) → Not real!
Oh no! So even though algebra gave us k=4/3, when we plug in, we get square root of negative number → not allowed in real numbers.
So no real solution.
✔ Final Answer: No real solution
---
10. m + 4 = √(m + 4)
Let u = √(m + 4), so u ≥ 0, and u² = m + 4 → m = u² - 4
Then equation becomes:
(u² - 4) + 4 = u → u² = u → u² - u = 0 → u(u - 1) = 0
So u = 0 or u = 1
If u = 0 → √(m+4)=0 → m+4=0 → m=-4
Check: -4 + 4 = 0; √(-4+4)=√0=0 → ✔
If u = 1 → √(m+4)=1 → m+4=1 → m=-3
Check: -3 + 4 = 1; √(-3+4)=√1=1 → ✔
Both work!
✔ Final Answer: m = -4 or m = -3
---
11. √(a - 3) - 1 = 4
Add 1:
√(a - 3) = 5
Square:
a - 3 = 25
a = 28
Check: √(28-3) -1 = √25 -1 = 5-1=4 → ✔
✔ Final Answer: a = 28
---
12. √(n² + 2) = n
Square both sides:
n² + 2 = n²
Subtract n²:
2 = 0 → ✘ Contradiction!
Also, note: √(n² + 2) > |n| ≥ n, so can never equal n unless... never.
✔ Final Answer: No solution
---
13. 3 - √(3b + 2) = 5
Subtract 3:
-√(3b + 2) = 2
Multiply both sides by -1:
√(3b + 2) = -2
Again, square root can’t be negative → no solution.
✔ Final Answer: No solution
---
14. √(c² + 9) = √(c² + 2c)
Square both sides:
c² + 9 = c² + 2c
Subtract c²:
9 = 2c
c = 9/2 = 4.5
Check:
Left: √((4.5)² + 9) = √(20.25 + 9) = √29.25
Right: √((4.5)² + 2*4.5) = √(20.25 + 9) = √29.25 → ✔
Works!
✔ Final Answer: c = 9/2
---
15. √(d - 2) = √(d - 4)
Square both sides:
d - 2 = d - 4
Subtract d:
-2 = -4 → ✘ Contradiction!
No solution.
✔ Final Answer: No solution
---
Final Answers Summary:
1. x = ±√21
2. No solution
3. y = 43/9
4. No solution
5. No solution
6. No solution
7. t = 1
8. No real solution
9. No real solution
10. m = -4 or m = -3
11. a = 28
12. No solution
13. No solution
14. c = 9/2
15. No solution
──────────────────────────────────────
Final Answer:
1. x = √21 or x = -√21
2. No solution
3. y = 43/9
4. No solution
5. No solution
6. No solution
7. t = 1
8. No real solution
9. No real solution
10. m = -4 or m = -3
11. a = 28
12. No solution
13. No solution
14. c = 9/2
15. No solution
---
1. √(x² + 4) = 5
Square both sides:
x² + 4 = 25
Subtract 4:
x² = 21
Take square root:
x = ±√21
Check:
Plug back into original: √((±√21)² + 4) = √(21 + 4) = √25 = 5 → ✔ Works for both.
✔ Final Answer: x = √21 or x = -√21
---
2. √(8x - 1) + 9 = 0
Isolate radical:
√(8x - 1) = -9
But wait — square roots are always ≥ 0 (in real numbers). You can’t have a square root equal to -9. So no solution.
✔ Final Answer: No solution
---
3. √(9y - 7) - 1 = 5
Add 1 to both sides:
√(9y - 7) = 6
Square both sides:
9y - 7 = 36
Add 7:
9y = 43
Divide by 9:
y = 43/9
Check:
√(9*(43/9) - 7) - 1 = √(43 - 7) - 1 = √36 - 1 = 6 - 1 = 5 → ✔
✔ Final Answer: y = 43/9
---
4. √(3x) = √(3x - 1)
Square both sides:
3x = 3x - 1
Subtract 3x from both sides:
0 = -1 → ✘ Contradiction!
No value of x makes this true.
✔ Final Answer: No solution
---
5. √(t + 1) = √(t - 1)
Square both sides:
t + 1 = t - 1
Subtract t:
1 = -1 → ✘ Contradiction!
No solution.
✔ Final Answer: No solution
---
6. x = √(x² + 4)
Square both sides:
x² = x² + 4
Subtract x²:
0 = 4 → ✘ Contradiction!
Wait — but let’s think: if x is positive, maybe? Try plugging in values.
Try x = 2: √(4 + 4) = √8 ≈ 2.8 ≠ 2
Try x = 3: √(9+4)=√13≈3.6≠3
Actually, since √(x² + 4) > |x| for all real x, it can never equal x unless... never.
So no solution.
✔ Final Answer: No solution
---
7. 4 + √(2t - 2) = 4
Subtract 4:
√(2t - 2) = 0
Square both sides:
2t - 2 = 0
2t = 2
t = 1
Check: 4 + √(2*1 - 2) = 4 + √0 = 4 → ✔
✔ Final Answer: t = 1
---
8. z + 4 = √(3z + 4)
Isolate radical? It’s already on one side. Square both sides:
(z + 4)² = 3z + 4
Expand left: z² + 8z + 16 = 3z + 4
Bring all to left: z² + 5z + 12 = 0
Discriminant: 25 - 48 = -23 → No real solutions.
Wait — did I do that right?
Left: (z+4)^2 = z² + 8z + 16
Right: 3z + 4
So: z² + 8z + 16 - 3z - 4 = 0 → z² + 5z + 12 = 0 → yes.
Discriminant negative → no real solution.
But let’s test small values just in case.
Try z=0: 0+4=4; √(0+4)=2 → 4≠2
z=1: 5 vs √7≈2.6 → no
z=-1: 3 vs √1=1 → no
z=-3: 1 vs √(-9+4)=√(-5) → invalid
So indeed, no real solution.
✔ Final Answer: No real solution
---
9. √(9 - 7k) = √(1 - k)
Square both sides:
9 - 7k = 1 - k
Add 7k to both sides:
9 = 1 + 6k
Subtract 1:
8 = 6k
k = 8/6 = 4/3
Check:
Left: √(9 - 7*(4/3)) = √(9 - 28/3) = √(27/3 - 28/3) = √(-1/3) → Not real!
Oh no! So even though algebra gave us k=4/3, when we plug in, we get square root of negative number → not allowed in real numbers.
So no real solution.
✔ Final Answer: No real solution
---
10. m + 4 = √(m + 4)
Let u = √(m + 4), so u ≥ 0, and u² = m + 4 → m = u² - 4
Then equation becomes:
(u² - 4) + 4 = u → u² = u → u² - u = 0 → u(u - 1) = 0
So u = 0 or u = 1
If u = 0 → √(m+4)=0 → m+4=0 → m=-4
Check: -4 + 4 = 0; √(-4+4)=√0=0 → ✔
If u = 1 → √(m+4)=1 → m+4=1 → m=-3
Check: -3 + 4 = 1; √(-3+4)=√1=1 → ✔
Both work!
✔ Final Answer: m = -4 or m = -3
---
11. √(a - 3) - 1 = 4
Add 1:
√(a - 3) = 5
Square:
a - 3 = 25
a = 28
Check: √(28-3) -1 = √25 -1 = 5-1=4 → ✔
✔ Final Answer: a = 28
---
12. √(n² + 2) = n
Square both sides:
n² + 2 = n²
Subtract n²:
2 = 0 → ✘ Contradiction!
Also, note: √(n² + 2) > |n| ≥ n, so can never equal n unless... never.
✔ Final Answer: No solution
---
13. 3 - √(3b + 2) = 5
Subtract 3:
-√(3b + 2) = 2
Multiply both sides by -1:
√(3b + 2) = -2
Again, square root can’t be negative → no solution.
✔ Final Answer: No solution
---
14. √(c² + 9) = √(c² + 2c)
Square both sides:
c² + 9 = c² + 2c
Subtract c²:
9 = 2c
c = 9/2 = 4.5
Check:
Left: √((4.5)² + 9) = √(20.25 + 9) = √29.25
Right: √((4.5)² + 2*4.5) = √(20.25 + 9) = √29.25 → ✔
Works!
✔ Final Answer: c = 9/2
---
15. √(d - 2) = √(d - 4)
Square both sides:
d - 2 = d - 4
Subtract d:
-2 = -4 → ✘ Contradiction!
No solution.
✔ Final Answer: No solution
---
Final Answers Summary:
1. x = ±√21
2. No solution
3. y = 43/9
4. No solution
5. No solution
6. No solution
7. t = 1
8. No real solution
9. No real solution
10. m = -4 or m = -3
11. a = 28
12. No solution
13. No solution
14. c = 9/2
15. No solution
──────────────────────────────────────
Final Answer:
1. x = √21 or x = -√21
2. No solution
3. y = 43/9
4. No solution
5. No solution
6. No solution
7. t = 1
8. No real solution
9. No real solution
10. m = -4 or m = -3
11. a = 28
12. No solution
13. No solution
14. c = 9/2
15. No solution
Parent Tip: Review the logic above to help your child master the concept of solving square root equations worksheet.