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Zorluk: OrtaRational and Radical Expressions and Equations

What is the real solution to the equation 3x+1=x−3\sqrt{3x + 1} = x - 3?

  1. A
    1
  2. B
    1 and 8
  3. 8Cevap
  4. D
    -2 and 5
  5. E
    -2

Cevap

The only real solution is 8.
The correct answer is the single value 8. Squaring both sides of the equation 3x+1=x−3\sqrt{3x+1} = x-3 leads to the quadratic equation x2−9x+8=0x^2-9x+8=0, which factors as (x−8)(x−1)=0(x-8)(x-1)=0. This yields potential solutions of 8 and 1. Checking 8 in the original equation gives 25=5\sqrt{25} = 5, which is correct. Checking 1 in the original equation gives 4=−2\sqrt{4} = -2, which is incorrect because the principal square root must be non-negative. Therefore, 1 is an extraneous solution, and 8 is the only valid solution.

Adım Adım Çözüm

1
Square both sides of the equation to eliminate the radical.
3x+1=(x−3)23x + 1 = (x - 3)^2
To clear the square root and obtain a polynomial equation.
2
Expand the right side and move all terms to the right side to set the quadratic equation to zero.
x2−9x+8=0x^2 - 9x + 8 = 0
Expanding (x−3)2(x - 3)^2 gives x2−6x+9x^2 - 6x + 9. Subtracting 3x3x and 11 from both sides gives the standard form of the quadratic equation.
3
Factor the quadratic equation.
(x−8)(x−1)=0(x - 8)(x - 1) = 0
Finding two numbers that multiply to 8 and add to -9, which are -8 and -1, allows us to find potential solutions x=8x = 8 and x=1x = 1.
4
Check both potential solutions in the original equation to identify extraneous solutions.
Checking x=8x = 8: 3(8)+1=25=5\sqrt{3(8) + 1} = \sqrt{25} = 5 and 8−3=58 - 3 = 5 (True). Checking x=1x = 1: 3(1)+1=4=2\sqrt{3(1) + 1} = \sqrt{4} = 2 and 1−3=−21 - 3 = -2 (False).
Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original equation.

Anahtar Kavram

Solving radical equations and identifying extraneous solutions
Tahmini Süre:1m 30s
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Konular

Rational and Radical Expressions and Equations

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