Multiple choice

The equation $x + \sqrt {x - 2} = 4$ has

  1. $2$ real roots
  2. $1$ real and $1$ imaginary root
  3. $2$ imaginary roots

  4. No roots
  5. $1$ real root
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Let sqrt(x-2) = y, so x-2 = y^2 or x = y^2 + 2. Substituting into the equation: y^2 + 2 + y = 4, which is y^2 + y - 2 = 0. Factoring gives (y+2)(y-1) = 0, so y = 1 or y = -2. Since y = sqrt(x-2) must be non-negative, y = 1. Thus, sqrt(x-2) = 1, x-2 = 1, x = 3. Checking x=3: 3 + sqrt(3-2) = 3+1 = 4. There is only one real root.

AI explanation

Isolate the radical to get sqrt(x - 2) = 4 - x, then square both sides to obtain the quadratic equation x - 2 = (4 - x)^2. Expanding and rearranging gives x^2 - 9x + 18 = 0, which factors into (x - 3)(x - 6) = 0, yielding potential roots of 3 and 6. Checking these in the original equation x + sqrt(x - 2) = 4, we test x = 3 to get 3 + sqrt(1) = 4, which is valid. Testing x = 6 gives 6 + sqrt(4) = 4, which is false, leaving only one real root of 3.

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