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PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Which statement best reflects the solution(s) of the equation?

PLEASE HELP ASAP CORRECT ANSWERS ONLY PLEASE I CANNOT RETAKE THIS Which statement best reflects the solutions of the equation class=

Respuesta :

Rinkhals

[tex]\sqrt{3x + 1} = x - 3\\\\3x + 1 = (x - 3)^2\\\\3x + 1 = x^2 + 9 - 6x\\\\x^2 - 9x + 8 = 0\\\\x = 8\;or\;x = 1[/tex]

but x = 1 does not satisfy the Equation

So, The answer is :

There is only one solution : x = 8

The solution x = 1 is an extraneous solution

tramserran

Answer: x = 8

              x = 1 is an extraneous solution

Step-by-step explanation:

[tex]\sqrt{3x+1}[/tex] - x + 3 = 0

RESTRICTIONS: 3x + 1 ≥ 0   →   x ≥ [tex]-\frac{1}{3}[/tex]  (because square roots cannot be less than zero/negative)

[tex]\sqrt{3x+1}[/tex] - x + 3 = 0

             +x  -3 +x  -3

[tex]\sqrt{3x+1}[/tex]              = x - 3

[tex](\sqrt{3x+1})^{2}[/tex]           = (x - 3)²

3x + 1                 = x² - 6x + 9

-3x   -1                          -3x   -1

                      0  = x² - 9x + 8

                      0  = (x - 1)(x - 8)

                 0 = x - 1         0 = x - 8

                +1      +1         +8     +8

                  1 = x              8 = x      both solutions are ≥ [tex]-\frac{1}{3}[/tex]

Check:  

x = 1: [tex]\sqrt{3(1)+1}[/tex] - (1) + 3 = 0

         2 - 1 + 3 = 0

                    4 = 0

                   FALSE   so x = 1 is NOT a valid solution

x = 8: [tex]\sqrt{3(8)+1}[/tex] - (8) + 3 = 0

         5 - 8 + 3 = 0

                    0 = 0

                    TRUE   so x = 8 is a valid solution

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