Exercise 74 Page 644

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. 2t^2-t-28=0 ⇕ 2t^2+( - 1)t+( -28)=0We can see that a= 2, b= - 1, and c= -28. Let's substitute these values into the Quadratic Formula.

t=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

t=- ( -1)±sqrt(( - 1)^2-4( 2)( -28))/2( 2)

▼

Evaluate right-hand side

NegNeg

- (- a)=a

t=1±sqrt((- 1)^2-4(2)(-28))/2(2)

NegBaseToPosBase

(- a)^2=a^2

t=1±sqrt(1-4(2)(-28))/2(2)

Multiply

Multiply

t=1±sqrt(1-8(-28))/4

MultNegNegOnePar

- a(- b)=a* b

t=1±sqrt(1+224)/4

AddTerms

Add terms

t=1±sqrt(225)/4

CalcRoot

Calculate root

t=1± 15/4

Using the Quadratic Formula, we found that the solutions of the given equation are t= 1± 15 4.

t=1± 15/4
t_1=1+15/4	t_2=1-15/4
t_1=16/4	t_2=-14/4
t_1= 4	t_2= -7/2

Therefore, the solutions are t_1=4 and t_2=- 72. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check t_1=4 and t_2=- 72 one at a time.

t_1=4

Let's substitute t=4 into the original equation.

2t=sqrt(2t+56)

Substitute

t= 4

2( 4)? =sqrt(2( 4)+56)

▼

Simplify

Multiply

Multiply

8? =sqrt(8+56)

AddTerms

Add terms

8? =sqrt(64)

CalcRoot

Calculate root

8=8 ✓

In this case we got a true statement. Therefore, t=4 is a solution of the original equation.

t_2=- 72

Now, let's substitute t= - 72.

2t=sqrt(2t+56)

Substitute

t= -7/2

2( -7/2)? =sqrt(2( -7/2)+56)

▼

Simplify

MultPosNeg

a(- b)=- a * b

-2 * 7/2? =sqrt(- 2* 7/2+56)

DenomMultFracToNumber

A * a/A= a

-7? =sqrt(-7+56)

AddTerms

Add terms

-7? =sqrt(49)

CalcRoot

Calculate root

-7 ≠ 7 *

We got a false statement, so t=- 72 is an extraneous solution. Therefore, t=4 is the only solution of the original equation.