Exercise 6 Page 644

We will find and check the solutions of the given equation.

Finding the Solutions

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! 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 -2n -3 = 0 ⇕ 1n^2+( - 2)n+( -3)=0 We can see that a= 1, b= - 2, and c= -3. Let's substitute these values into the Quadratic Formula.

n=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

n=- ( -2)±sqrt(( - 2)^2-4( 1)( -3))/2( 1)

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Solve for n and Simplify

NegNeg

- (- a)=a

n=2±sqrt((- 2)^2-4(1)(-3))/2(1)

NegBaseToPosBase

(- a)^2=a^2

n=2±sqrt(4-4(1)(-3))/2(1)

IdPropMult

Identity Property of Multiplication

n=2±sqrt(4-4(-3))/2

MultPosNeg

a(- b)=- a * b

n=2±sqrt(4-(-12))/2

SubNeg

a-(- b)=a+b

n=2±sqrt(4+12)/2

AddTerms

Add terms

n=2±sqrt(16)/2

CalcRoot

Calculate root

n=2± 4/2

Using the Quadratic Formula, we found that the solutions of the given equation are n= 2± 42.

n=2± 4/2
n_1=2+4/2	n_2=2-4/2
n_1=6/2	n_2=-2/2
n_1= 3	n_2= -1

Therefore, the solutions are n_1= 3 and n_2= -1. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check n_1=3 and n_2=-1 one at a time.

n_1=3

Let's substitute n= 3 into the original equation.

sqrt(2n+3)=n

Substitute

n= 3

sqrt(2( 3)+3)? = 3

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Simplify

Multiply

Multiply

sqrt(6+3)? =3

AddTerms

Add terms

sqrt(9)? =3

CalcRoot

Calculate root

3=3 ✓

We got a true statement, so n=3 is a solution.

n_2=-1

Now, let's substitute n= -1.

sqrt(2n+3)=n

Substitute

n= -1

sqrt(2( -1)+3)? = -1

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Simplify

MultPosNeg

a(- b)=- a * b

sqrt(-2+3)? = -1

AddTerms

Add terms

sqrt(1)? = -1

CalcRoot

Calculate root

1 ≠ -1 *

In this case we got a false statement. Therefore, n=3 is the only solution to the original equation.