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Zorluk: OrtaRadical and Rational Equations

What is the sum of all real values of xx that satisfy the equation 3x+40−x=4\sqrt{3x + 40} - x = 4?

  1. A
    -5
  2. 3Cevap
  3. C
    5
  4. D
    -8

Cevap

The sum of all valid real solutions is 3.
Isolating the radical gives 3x+40=x+4\sqrt{3x + 40} = x + 4. Squaring both sides and simplifying results in x2+5x−24=0x^2 + 5x - 24 = 0. Factoring gives (x+8)(x−3)=0(x + 8)(x - 3) = 0, leading to candidate solutions of 3 and -8. Substituting these back shows that only 3 satisfies the original equation, as -8 results in an extraneous solution. Therefore, the sum of all valid solutions is 3.

Adım Adım Çözüm

1
Isolate the radical expression on one side of the equation.
3x+40=x+4\sqrt{3x + 40} = x + 4
To prepare both sides of the equation for squaring.
2
Square both sides of the equation to eliminate the radical.
3x+40=(x+4)23x + 40 = (x + 4)^2
Squaring a square root cancels the radical.
3
Expand the right side and move all terms to one side to form a quadratic equation.
x2+5x−24=0x^2 + 5x - 24 = 0
Expanding (x+4)2(x+4)^2 yields x2+8x+16x^2 + 8x + 16. Subtracting 3x3x and 4040 from both sides sets the quadratic equation to zero.
4
Factor the quadratic equation.
(x+8)(x−3)=0(x + 8)(x - 3) = 0
The factors of -24 that add up to 5 are 8 and -3.
5
Solve for the potential values of xx.
x=−8x = -8 or x=3x = 3
Setting each factor to zero yields the candidate solutions.
6
Check each candidate solution in the original equation to identify any extraneous solutions.
For x=3x = 3, 3(3)+40−3=7−3=4\sqrt{3(3)+40}-3 = 7-3 = 4 (valid). For x=−8x = -8, 3(−8)+40−(−8)=4+8=12≠4\sqrt{3(-8)+40}-(-8) = 4+8 = 12 \neq 4 (extraneous).
Squaring both sides can introduce extraneous solutions that do not satisfy the original equation.

Anahtar Kavram

Solving radical equations and checking for extraneous solutions.
Bu soruyu puanla

Konular

Radical and Rational Equations

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