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Pearson Algebra 1 Common Core, 2011 View details

4. Solving Radical Equations

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Exercise 34 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!

2y=sqrt(5y+6)

RaiseEqn

LHS^2=RHS^2

(2y)^2=(sqrt(5y+6))^2

PowSqrt

( sqrt(a) )^2 = a

(2y)^2 =5y+6

PowProdII

(a b)^m=a^m b^m

2^2y^2=5y+6

CalcPow

Calculate power

4y^2=5y+6

▼

LHS-(5y+6)=RHS-(5y+6)

SubEqn

LHS-5y=RHS-5y

4y^2-5y=6

SubEqn

LHS-6=RHS-6

4y^2-5y-6=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. 4y^2-5y-6=0 ⇕ 4y^2+( - 5)y+( -6)=0

We can see that a= 4, b= - 5, and c= -6. Let's substitute these values into the Quadratic Formula.

y=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

y=- ( -5)±sqrt(( - 5)^2-4( 4)( -6))/2( 4)

▼

Evaluate

NegNeg

- (- a)=a

y=5±sqrt((- 5)^2-4(4)(-6))/2(4)

NegBaseToPosBase

(- a)^2=a^2