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Question

Difficulty: EasyRadical and Rational Equations

What is the solution to the equation 3x+13=x+3\sqrt{3x + 13} = x + 3?

  1. 1Answer
  2. B
    -4
  3. C
    1 and -4
  4. D
    4

Answer

1
The value 11 is the correct answer because substituting it back into the original equation results in a true mathematical statement: 3(1)+13=1+3\sqrt{3(1) + 13} = 1 + 3, which simplifies to 4=44 = 4.

Step-by-Step Solution

1
Square both sides of the equation to eliminate the radical.
3x+13=(x+3)23x + 13 = (x + 3)^2
Squaring both sides is the standard method to solve equations containing square roots.
2
Expand the right side of the equation.
3x+13=x2+6x+93x + 13 = x^2 + 6x + 9
Expanding the squared binomial gives a quadratic expression.
3
Move all terms to one side to set the quadratic equation to zero.
x2+3x−4=0x^2 + 3x - 4 = 0
Setting the quadratic equation to zero allows us to solve for xx by factoring.
4
Factor the quadratic equation.
(x+4)(x−1)=0(x + 4)(x - 1) = 0
Finding two numbers that multiply to −4-4 and add to 33 gives the binomial factors.
5
Solve for the potential values of xx.
x=−4x = -4 or x=1x = 1
Setting each factor equal to zero yields the potential solutions.
6
Substitute each potential solution back into the original equation to check for extraneous solutions.
For x=1x = 1: 3(1)+13=1+3⇒4=4\sqrt{3(1) + 13} = 1 + 3 \Rightarrow 4 = 4 (valid). For x=−4x = -4: 3(−4)+13=−4+3⇒1=−1\sqrt{3(-4) + 13} = -4 + 3 \Rightarrow 1 = -1 (invalid). Thus, the only solution is 11.
Squaring both sides can introduce extraneous solutions, so checking all potential solutions in the original equation is necessary to verify their validity.

Key Concept

Solving radical equations by squaring both sides and checking for extraneous solutions.
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Topics

Radical and Rational Equations

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