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Math  /  Algebra

QuestionQuestion
Solve for all possible values of x . 3x+18=x+6\sqrt{3 x+18}=x+6
Answer Attempt 1 out of 2 (†) Additional Solution No Solution

Studdy Solution

STEP 1

1. The equation involves a square root, which means we will need to square both sides to eliminate the square root.
2. We must check for extraneous solutions, as squaring both sides can introduce solutions that do not satisfy the original equation.

STEP 2

1. Isolate the square root and square both sides.
2. Simplify and solve the resulting quadratic equation.
3. Verify solutions in the original equation to check for extraneous solutions.

STEP 3

The equation is already isolated with the square root on one side:
3x+18=x+6 \sqrt{3x + 18} = x + 6
Square both sides to eliminate the square root:
(3x+18)2=(x+6)2 (\sqrt{3x + 18})^2 = (x + 6)^2 3x+18=(x+6)2 3x + 18 = (x + 6)^2

STEP 4

Expand the right side of the equation:
3x+18=x2+12x+36 3x + 18 = x^2 + 12x + 36
Rearrange the equation to set it to zero:
x2+12x+36−3x−18=0 x^2 + 12x + 36 - 3x - 18 = 0 x2+9x+18=0 x^2 + 9x + 18 = 0

STEP 5

Solve the quadratic equation using the quadratic formula, x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=1 a = 1 , b=9 b = 9 , and c=18 c = 18 :
Calculate the discriminant:
b2−4ac=92−4⋅1⋅18 b^2 - 4ac = 9^2 - 4 \cdot 1 \cdot 18 =81−72 = 81 - 72 =9 = 9
Since the discriminant is positive, there are two real solutions:
x=−9±92⋅1 x = \frac{-9 \pm \sqrt{9}}{2 \cdot 1} x=−9±32 x = \frac{-9 \pm 3}{2}
Calculate the two possible solutions:
x1=−9+32=−62=−3 x_1 = \frac{-9 + 3}{2} = \frac{-6}{2} = -3 x2=−9−32=−122=−6 x_2 = \frac{-9 - 3}{2} = \frac{-12}{2} = -6

STEP 6

Verify each solution in the original equation to check for extraneous solutions.
For x=−3 x = -3 :
3(−3)+18=−3+6 \sqrt{3(-3) + 18} = -3 + 6 −9+18=3 \sqrt{-9 + 18} = 3 9=3 \sqrt{9} = 3 3=3 3 = 3
This solution is valid.
For x=−6 x = -6 :
3(−6)+18=−6+6 \sqrt{3(-6) + 18} = -6 + 6 −18+18=0 \sqrt{-18 + 18} = 0 0=0 \sqrt{0} = 0 0=0 0 = 0
This solution is also valid.
The possible values of x x are:
−3and−6 \boxed{-3} \quad \text{and} \quad \boxed{-6}

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