Multiple choice

The equation $y+\sqrt{y-4}=6$ has

  1. two real roots

  2. only one real root

  3. no real roots

  4. one real root and one imaginary root

  5. only one imaginary root
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let sqrt(y-4) = x, so y = x^2 + 4. Substituting into the equation: x^2 + 4 + x = 6, which is x^2 + x - 2 = 0. Factoring gives (x+2)(x-1) = 0. Since x must be non-negative (as it is a square root), x = 1. Thus, sqrt(y-4) = 1, so y-4 = 1, y = 5. Only one real root exists.

AI explanation

Isolating the radical gives the square root of (y - 4) = 6 - y. Squaring both sides results in y - 4 = 36 - 12y + y^2, which simplifies to the quadratic equation y^2 - 13y + 40 = 0. Factoring provides potential roots of 5 and 8, but substituting y = 8 back into the original equation results in the false statement 10 = 6. Only y = 5 satisfies the original equation, meaning it has exactly one real root.

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