Multiple choice

The number of roots of the equation $9+\sqrt{x-3}=x$, is

  1. $0$
  2. $1$
  3. $2$
  4. $3$
  5. $\infty$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let sqrt(x-3) = y, then x = y^2+3. Equation: 9+y = y^2+3 => y^2-y-6=0. (y-3)(y+2)=0. y=3 or y=-2. Since y must be >=0, y=3. Then sqrt(x-3)=3 => x-3=9 => x=12. Only one root.

AI explanation

To solve the radical equation 9 + sqrt(x-3) = x, first isolate the radical to get sqrt(x-3) = x - 9. Squaring both sides gives x - 3 = x^2 - 18x + 81, which rearranges to the quadratic equation x^2 - 19x + 84 = 0. Factoring this equation yields (x - 7)(x - 12) = 0, giving potential roots of 7 and 12. Checking these in the original equation, x = 12 works because sqrt(12-3) = 12 - 9 evaluates to 3 = 3, but x = 7 fails because sqrt(7-3) = 7 - 9 evaluates to 2 = -2, which is false. Therefore, the equation has exactly 1 valid root.

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