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Pearson Algebra 1 Common Core, 2011 View details

4. Solving Radical Equations

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Exercise 32 Page 637

Raise both sides of the radical equation to a power equal to the index of the radical.

Practice makes perfect

To solve equations with a variable expression inside a radical, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation using this method!

n=sqrt(4n+5)

RaiseEqn

LHS^2=RHS^2

n^2=(sqrt(4n+5))^2

PowSqrt

( sqrt(a) )^2 = a

n^2 =4n+5

▼

LHS-(4n+5)=RHS-(4n+5)

SubEqn

LHS-4n=RHS-4n

n^2-4n=5

SubEqn

LHS-5=RHS-5

n^2-4n-5=0

We now have a quadratic equation, and we need to find its roots. To do it, let's identify the values of a, b, and c. n^2-4n-5=0 ⇕ 1n^2+( - 4)n+( -5)=0

We can see that a= 1, b= - 4, and c= -5. Let's substitute these values into the Quadratic Formula.

n=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

n=- ( -4)±sqrt(( - 4)^2-4( 1)( -5))/2( 1)

▼

Evaluate

NegNeg

- (- a)=a

n=4±sqrt((- 4)^2-4(1)(-5))/2(1)

NegBaseToPosBase

(- a)^2=a^2