Grade 9 · Algebra & functions

Radical word problems (ACT) practice

Radical word problems (ACT) is a grade 9 math skill aligned to Common Core standard HSN.RN.A.2: rewrite expressions involving radicals and rational exponents using the properties of exponents. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 radical word problems (act) problems our math games drill.

CCSS HSN.RN.A.210 questions in the bank
Sample questions

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Question 1easy

The speed, in meters per second, of a certain water wave can be estimated by multiplying the wavelength, in meters, by 1212 and then taking the square root of the product. According to this method, what is the estimated speed, in meters per second, of a wave with a wavelength of 4545 meters?

The product is 45⋅12=54045 \cdot 12 = 540. Factor 540=36⋅15540 = 36 \cdot 15, where 3636 is the largest perfect-square factor. So 540=36⋅15=615\sqrt{540} = \sqrt{36 \cdot 15} = 6\sqrt{15}, and 15=3⋅515 = 3 \cdot 5 is square-free.

Question 2easy

The sound intensity level, in a simplified unit, of a certain speaker can be estimated by multiplying the speaker's power, in watts, by 88 and then taking the square root of the product. According to this method, what is the estimated intensity level of a speaker with a power of 7575 watts?

The product is 75⋅8=60075 \cdot 8 = 600. Factor 600=100⋅6600 = 100 \cdot 6, where 100100 is the largest perfect-square factor. So 600=100⋅6=106\sqrt{600} = \sqrt{100 \cdot 6} = 10\sqrt{6}, and 6=2⋅36 = 2 \cdot 3 is square-free.

Question 3medium

The impact speed, in feet per second, of an object dropped from a height can be estimated by multiplying the height, in feet, by 6464 and then taking the square root of the product. According to this method, what is the estimated impact speed, in feet per second, of an object dropped from a height of 1212 feet?

The product is 12⋅64=76812 \cdot 64 = 768. Factor 768=256⋅3768 = 256 \cdot 3, where 256256 is the largest perfect-square factor. So 768=256⋅3=163\sqrt{768} = \sqrt{256 \cdot 3} = 16\sqrt{3}, and 33 is square-free.

Question 4medium

The diagonal length, in inches, of a certain square television screen can be estimated by multiplying the screen's area, in square inches, by 22 and then taking the square root of the product. According to this method, what is the estimated diagonal length, in inches, of a screen with an area of 216216 square inches?

The product is 216⋅2=432216 \cdot 2 = 432. Factor 432=144⋅3432 = 144 \cdot 3, where 144144 is the largest perfect-square factor. So 432=144⋅3=123\sqrt{432} = \sqrt{144 \cdot 3} = 12\sqrt{3}, and 33 is square-free.

Question 5medium

An estimate of escape speed, in kilometers per second, from a small asteroid can be found by multiplying the asteroid's radius, in kilometers, by 88 and then taking the square root of the product. According to this method, what is the estimated escape speed, in kilometers per second, from an asteroid with a radius of 112112 kilometers?

The product is 112⋅8=896112 \cdot 8 = 896. Factor 896=64⋅14896 = 64 \cdot 14, where 6464 is the largest perfect-square factor. So 896=64⋅14=814\sqrt{896} = \sqrt{64 \cdot 14} = 8\sqrt{14}, and 14=2⋅714 = 2 \cdot 7 is square-free.

Question 6medium

The coverage distance, in feet, of a certain spray paint can be estimated by multiplying the number of ounces of paint by 4848 and then taking the square root of the product. According to this method, what is the estimated coverage distance, in feet, when 1818 ounces of paint are used?

The product is 18⋅48=86418 \cdot 48 = 864. Factor 864=144⋅6864 = 144 \cdot 6, where 144144 is the largest perfect-square factor. So 864=144⋅6=126\sqrt{864} = \sqrt{144 \cdot 6} = 12\sqrt{6}, and 6=2⋅36 = 2 \cdot 3 is square-free.

Question 7medium

The tension, in newtons, in a certain cable can be estimated by multiplying the length of the cable, in meters, by 2828 and then taking the square root of the product. According to this method, what is the estimated tension, in newtons, in a cable that is 4545 meters long?

The product is 45⋅28=126045 \cdot 28 = 1260. Factor 1260=36⋅351260 = 36 \cdot 35, where 3636 is the largest perfect-square factor. So 1260=36⋅35=635\sqrt{1260} = \sqrt{36 \cdot 35} = 6\sqrt{35}, and 35=5⋅735 = 5 \cdot 7 is square-free.

Question 8medium

The safe descent speed, in miles per hour, on a certain ramp can be estimated by multiplying the height of the ramp, in feet, by 5050 and then taking the square root of the product. According to this method, what is the estimated safe descent speed, in miles per hour, on a ramp that is 2828 feet high?

The product is 28⋅50=140028 \cdot 50 = 1400. Factor 1400=100⋅141400 = 100 \cdot 14, where 100100 is the largest perfect-square factor. So 1400=100⋅14=1014\sqrt{1400} = \sqrt{100 \cdot 14} = 10\sqrt{14}, and 14=2⋅714 = 2 \cdot 7 is square-free.

Tests that cover radical word problems (act):
NAEPSATSBACNC EOG/EOCVTCAP
More Grade 9 skills:
Radical to exponent conversionsRational equation solving (SAT)Quadratic minimum x-valueRational exponentsQuadratic function word problems (ACT)Scalar times 2×2 matrix (ACT)Quadratic formulaSimplifying square roots (ACT)All Grade 9 practice →

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