Exercise 18 Page 261

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. In order to do that, we can factor the obtained equation. Let's factor out t from the whole expression. t^2+4t = 0 ⇔ t(t+4)=0 Now, we can use the Zero Product Property to determine the roots of our equation.

t(t+4)=0

ZeroProdProp

Use the Zero Product Property

lct=0 & (I) t+4=0 & (II)

SubEqn

(II): LHS-4=RHS-4

lt=0 t=-4

Therefore, the solutions are t_1= 0 and t_2= -4. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check t_1=0 and t_2=-4 one at a time.

t_1=0

Let's substitute t= 0 into the original equation.

sqrt(1-2t)=1+t

Substitute

t= 0

sqrt(1-2( 0))? =1+ 0

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Simplify

ZeroPropMult

Zero Property of Multiplication

sqrt(1-0)? =1+0

AddSubTerms

Add and subtract terms

sqrt(1)? = 1

CalcRoot

Calculate root

1=1 ✓

We got a true statement, so t=0 is a solution.

t_2=-4

Now, let's substitute t= -4.

sqrt(1-2t)=1+t

Substitute

t= -4

sqrt(1-2( -4))? =1+( -4)

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Simplify

MultPosNeg

a(- b)=- a * b

sqrt(1-(-8))? =1+(-4)

SubNeg

a-(- b)=a+b

sqrt(1+8)? =1+(-4)

AddNeg

a+(- b)=a-b

sqrt(1+8)? =1-4

AddSubTerms

Add and subtract terms

sqrt(9)? = -3

CalcRoot

Calculate root

3 ≠ -3 *

In this case we got a false statement. Therefore, t=0 is the only solution to the original equation.