Radicals Practice Questions

To review these concepts, go to Radicals Study Guide.

Radicals Practice Questions

Problems

Practice 1

Simplify each expression as much as possible.

1. √m10
2. 
3. ³√j15
4. 2√25v2
5. –4³√27k6

Practice 2

Solve the following equations.

1. h³ = 64
2. c² – 8 = 1
3. 5n² = 80
4. √3p = 9
5. ³√5d = 5

Solutions

Practice 1

1. The square root of m¹⁰ is the quantity that, when multiplied by itself, is equal to m¹⁰. To find the square root of a base with an even exponent, divide the exponent by 2: 10 ÷ 2 = 5. (m⁵)(m⁵) = m¹⁰, which is why √m10 = m⁵.
2. The numerator of a fractional exponent is the power to which the variable is raised. The denominator of the fractional exponent is the root to take of the variable. is equal to³√c4, which is the third root of c to the fourth power. There is no whole exponent of c, such that we could multiply it by itself three times and arrive at c⁴, so this expression cannot be simplified any further.
3. The third root of j¹⁵ is the quantity that, when multiplied three times, is equal to j¹⁵. To find the third root of a base, divide the exponent of the base by 3: 15 ÷ 3 = 5. (j⁵)(j⁵)(j⁵) = j¹⁵, which is why ³√j15 = j⁵.
4. The square root of 25v² is the quantity that, when multiplied by itself, is equal to 25v². Find the square root of the coefficient and the square root of the base with its exponent. The square root of 25 is 5, because (5)(5) = 25. To find the square root of a base with an even exponent, divide the exponent by 2: 2 ÷ 2 = 1. (v)(v) = v², which is why √v2 = v. Therefore, √25v2 = 5v. Multiply the coefficient of the radical, 2, by 5v: 2(5v) = 10v.
5. The third root of 27k⁶ is the quantity that, when multiplied three times, is equal to 27k⁶. Find the third root of the coefficient and the third root of the base with its exponent. The third root of 27 is 3, because (3)(3)(3) = 27. To find the third root of a base, divide the exponent of the base by 3: 6 ÷ 3 = 2. (k²)(k²)(k²) = k⁶, which is why ³√k6 = k². Therefore,³√27k6 = 3k². Multiply the coefficient of the radical, –4, by 3k²: –4(3k²) = –12k².

Practice 2

1. In the equation h³ = 64, h is raised to the third power. To get h alone on the left side of the equation, we must take the third root of both sides of the equation. The third root of h³ is h, because (h)(h)(h) = h³. The third root of 64 is 4, because (4)(4)(4) = 64.

   ³√h3 = ³√64

   h = 4

2. In the equation c² – 8 = 1, c is raised to the second power and 8 is subtracted from that square. First, add 8 to both sides of the equation so that the variable is on one side of the equation and the constant is on the other side:

   c² – 8 + 8 = 1 + 8

   c² = 9

   To get c alone on the left side of the equation, we must take the square root of both sides of the equation. The square root of c² is c, because (c)(c) = c². Because we are taking the square root, an even root, of a constant on the right side of the equation, we must take the plus and minus square roots. The positive square root of 9 is 3, since (3)(3) = 9. The negative square root of 9 is –3, because (–3)(–3) = 9.

   √c2 = ±√9

   c = 3, –3

3. In the equation 5n² = 80, n is raised to the second power and then multiplied by 5. First, divide both sides of the equation by 5 so that the variable is on one side of the equation and the constant is on the other side:

   

   n² = 16

   To get n alone on the left side of the equation, we must take the square root of both sides of the equation. The square root of n² is n, because (n)(n) = n². Because we are taking the square root, an even root, of a constant on the right side of the equation, we must take the plus and minus square roots. The positive square root of 16 is 4, because (4)(4) = 16. The negative of the square root of 16 is –4, because (–4)(–4) = 16.

   √n2 = ±√16

   n = 4, –4

4. In the equation √3p = 9, the square root of 3p is equal to 9. To remove the radical symbol from the left side of the equation, we must raise both sides of the equation to the second power.

   (√3p)² = (9)²

   3p = 81

   Because p is multiplied by 3, divide both sides of the equation by 3:

   

   p = 27

5. In the equation ³√5d = 5, the third root of 5d is equal to 5. To remove the radical symbol from the left side of the equation, we must raise both sides of the equation to the third power.

   (³√5d)³ = (5)³

   5d = 125

   Because d is multiplied by 5, divide both sides of the equation by 5:

   

   d = 25