Exercise 55 Page &nbsp; 267

Calculate root

x=10 ± 6/2

From here, we break the solution into two cases and solve each. Positive:&& x_1=10+6/2 ⇔ x_1=8 [0.8em] Negative:&& x_2=10-6/2 ⇔ x_2=2 Squaring the equation, gives us a second degree equation with two solutions. Therefore, we have to investigate if any of them is extraneous by substituting them into the original equation and checking if the equation holds true.

x-4=sqrt(2x)

x= 2

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

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Simplify

2-4? =sqrt(4)

Calculate root

2-4? =2

Subtract term

- 2 ≠ 2

The solution x=2 is extraneous. Let's check the other solution.

x-4=sqrt(2x)

x= 8

8-4? =sqrt(2( 8))

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Simplify

8-4? =sqrt(16)

Calculate root

8-4? =4

Subtract term

4=4

The second solution x=8 is valid.

Solving the second equation

To solve the equation, we have to square both sides to eliminate the radicand. Note that the minus sign will disappear when you square the right-hand side.

x-4=- sqrt(2x)

RaiseEqn

LHS^2=RHS^2

(x-4)^2=2x

ExpandNegPerfectSquare

(a-b)^2=a^2-2ab+b^2

x^2-8x+16=2x

SubEqn

LHS-2x=RHS-2x

x^2-10x+16=0

Note that this is the same equation as we previously solved. Therefore, our solutions will still be x=2 and x=8. Like before, we have to test them in the original equation.

x-4=- sqrt(2x)

x= 2

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

▼

Simplify

2-4? =- sqrt(4)

Calculate root

2-4? =- 2

Subtract term

- 2 =- 2

Now, x=2, is a valid solution. Let's also check the second solution.

x-4=- sqrt(2x)

x= 8

8-4? =- sqrt(2( 8))

▼

Simplify

8-4? =- sqrt(16)

Calculate root

8-4? =- 4