It is given that the equation √x+p + √x = p has at least one real solution for x, where p is a real constant.
What is the complete set of possible values for p?
- Ap = 0 or p = 1
- Bp = 0 or p ≥ 1
- Cp ≥ −x
- Dp ≥ √x
- Ep ≥ 0
- Fp ≥ 1
Show the answer and worked solution
answer · B
- Ap = 0 or p = 1
- Bp = 0 or p ≥ 1
- Cp ≥ −x
- Dp ≥ √x
- Ep ≥ 0
- Fp ≥ 1
The answer must be a condition on p alone, which rules out the two options that still mention x. Both square roots are non-negative, so p ≥ 0 at once, and p = 0 forces √x = 0, so x = 0 works. For p > 0, isolate one root: √x+p = p − √x, and squaring gives x + p = p2 − 2p√x + x. Dividing by p leaves 1 = p − 2√x, so √x = p−12, which needs p ≥ 1; and that value automatically satisfies √x ≤ p, so the squaring introduced nothing false. Hence p = 0 or p ≥ 1.