Exercise 36 Page 870

We will solve the rational equation by cross multiplying. If a numerator or denominator contains addition or subtraction, be sure to treat each one as a parenthetical factor in the cross multiplication process. We obtained a quadratic equation. Let's identify the values of a, b, and c. 3x^2 - 3x - 18=0 ⇕ 3x^2 + ( -3)x + ( -18)=0 We see that a = 3, b = -3, and c = - 18. Next, we will substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

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Solve using the quadratic formula

SubstituteValues

Substitute values

x=- ( -3)±sqrt(( -3)^2-4( 3)( - 18))/2( 3)

NegNeg

- (- a)=a

x=3 ±sqrt((-3)^2-4(3)(- 18))/2(3)

CalcPow

Calculate power

x=3 ±sqrt(9-4(3)(- 18))/2(3)

Multiply

Multiply

x=3 ±sqrt(9+216)/6

AddTerms

Add terms

x=3 ±sqrt(225)/6

CalcRoot

Calculate root

x = 3 ± 15/6

We will find the values for x by using the positive and negative signs.

x = 3 ± 15/6
x = 3 + 15/6	x = 3 - 15/6
x=18/6	x=- 12/6
x=3	x=- 2

We found that x=3, and x=-2 are the possible values of x. Therefore, the answer is D.