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Difficulty: MediumRadical and Rational Equations

The equation 2x+12−x=−6\sqrt{2x + 12} - x = -6 has one real solution. What is this solution?

Answer: 12

Answer

12
Substituting 12 into the original equation 2x+12−x=−6\sqrt{2x + 12} - x = -6 yields 2(12)+12−12=36−12=6−12=−6\sqrt{2(12) + 12} - 12 = \sqrt{36} - 12 = 6 - 12 = -6. Since this creates a true statement, 12 is the unique real solution to the equation.

Step-by-Step Solution

1
Isolate the radical term by adding xx to both sides of the equation.
2x+12=x−6\sqrt{2x + 12} = x - 6
Isolating the radical on one side allows us to eliminate it by squaring both sides in the next step.
2
Square both sides of the equation to clear the square root.
2x+12=(x−6)22x + 12 = (x - 6)^2
Squaring a square root cancels the radical, allowing us to solve the equation algebraically.
3
Expand the squared binomial on the right-hand side.
2x+12=x2−12x+362x + 12 = x^2 - 12x + 36
Expanding (x−6)2(x - 6)^2 to x2−12x+36x^2 - 12x + 36 helps set up a standard quadratic form.
4
Subtract 2x2x and 1212 from both sides to form a quadratic equation equal to zero.
x2−14x+24=0x^2 - 14x + 24 = 0
A quadratic equation must be in the form ax2+bx+c=0ax^2 + bx + c = 0 to solve by factoring or the quadratic formula.
5
Factor the quadratic equation.
(x−12)(x−2)=0(x - 12)(x - 2) = 0
Factoring finds two numbers that multiply to 24 and add to -14, which are -12 and -2.
6
Test the potential solutions x=12x = 12 and x=2x = 2 in the original equation to identify any extraneous solutions.
For x=2x = 2: 2(2)+12−2=16−2=4−2=2≠−6\sqrt{2(2) + 12} - 2 = \sqrt{16} - 2 = 4 - 2 = 2 \neq -6. For x=12x = 12: 2(12)+12−12=36−12=6−12=−6\sqrt{2(12) + 12} - 12 = \sqrt{36} - 12 = 6 - 12 = -6. Thus, x=12x = 12 is the only valid solution.
Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original equation.

Key Concept

Solving radical equations by isolating the radical, squaring both sides, solving the resulting quadratic equation, and verifying all potential solutions to check for extraneous values.
Estimated Time:1m 30s
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Topics

Radical and Rational Equations

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