A student attempts to solve the following problem, where a and b are non-zero real numbers:
Show that if a2 − 4b3 ≥ 0 then there exist real numbers x and y such that a = xy(x+y) and b = xy.
Consider the following attempt:
(x−y)2 ≥ 0 (I)
so x2 + y2 − 2xy ≥ 0 (II)
so (x+y)2 − 4xy ≥ 0 (III)
so x2y2(x+y)2 − 4x3y3 ≥ 0 (IV)
so a2 − 4b3 ≥ 0 (V)
Which of the following best describes this attempt?
- AIt is completely correct.
- BIt is incorrect, but it would be correct if written in the reverse order.
- CIt is incorrect, but the student has correctly proved the converse.
- DIt is incorrect because there is an error in line (II).
- EIt is incorrect because there is an error in line (III).
- FIt is incorrect because there is an error in line (IV).
Show the answer and worked solution
answer · C
- AIt is completely correct.
- BIt is incorrect, but it would be correct if written in the reverse order.
- CIt is incorrect, but the student has correctly proved the converse.
- DIt is incorrect because there is an error in line (II).
- EIt is incorrect because there is an error in line (III).
- FIt is incorrect because there is an error in line (IV).
Check the lines first: (II) expands the square, (III) is the same expression since (x+y)2 − 4xy = x2 + y2 − 2xy, (IV) multiplies through by x2y2 ≥ 0, and (V) substitutes a = xy(x+y) and b = xy. Every line follows, so no individual line is wrong. What is wrong is the direction: the student starts from real x and y already existing and deduces a2 − 4b3 ≥ 0, which is the converse of the problem. Reversing the order would not fix it, since the argument would still assume x and y exist rather than producing them.