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McGraw Hill Integrated II, 2012 View details

3. Simplifying Radical Expressions

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Exercise 50 Page 249

Practice makes perfect

a We are given that the speed v of a ball can be determined by the following equation.

v=sqrt(2k/m) In this equation, k is the kinetic energy and m is the mass of the ball. We want to simplify this equation when the ball has a mass of 3 kilograms. Therefore, we will substitute 3 for m and rationalize the denominator using the Quotient Property and the Product Property of Square Roots.

v=sqrt(2k/m)

Substitute

m= 3

v=sqrt(2k/3)

SqrtQuot

sqrt(a/b)=sqrt(a)/sqrt(b)

v=sqrt(2k)/sqrt(3)

ExpandFrac

a/b=a * sqrt(3)/b * sqrt(3)

v=sqrt(2k)*sqrt(3)/sqrt(3)*sqrt(3)

ProdSqrt

sqrt(a)*sqrt(b)=sqrt(a* b)

v=sqrt(6k)/sqrt(3*3)

ProdToPowTwoFac

a* a=a^2

v=sqrt(6k)/sqrt(3^2)

SqrtPowToNumber

sqrt(a^2)=a

v=sqrt(6k)/3

b We are asked to determine the kinetic energy of the same ball when it is traveling 7 meters per second. Therefore, we need to substitute 7 for v into the simplified equation and solve it for k.

v=sqrt(6k)/3

Substitute

v= 7

7=sqrt(6k)/3

▼

Solve for k

MultEqn

LHS * 3=RHS* 3

21=sqrt(6k)

RaiseEqn

LHS^2=RHS^2

21^2=(sqrt(6k))^2