EPFO APFC 2023 General Studies · Question 75 of 120

Correct — C, (c) α, β < 0, that is, both roots are negative. Two lines of Vieta's relations settle it. For a monic quadratic x squared plus ax plus b equals 0, the sum of the roots is minus a and the product is b. Since b is given as positive, the product of the roots is positive, which means the two roots share a sign: either both are positive or both are negative. Since a is given as positive, the sum of the roots is negative, and two numbers of the same sign whose sum is negative must both be negative. The hypothesis that the roots are real is needed only so that talking about their signs makes sense; it amounts to the discriminant condition that a squared is at least 4b. There is an even quicker argument that avoids Vieta altogether. Take any x greater than or equal to zero. Then x squared is at least zero, ax is at least zero because a is positive, and b is strictly positive, so the whole expression is strictly positive and cannot be zero. No root can therefore be zero or positive, and since real roots exist by hypothesis, both must be negative. This second argument is worth internalising because it generalises: a polynomial with all coefficients positive can have no positive root, a fact usually met as part of Descartes' rule of signs. A concrete instance keeps the result honest. Take a equal to 5 and b equal to 4; the equation x squared plus 5x plus 4 equals 0 factorises as (x plus 1)(x plus 4) equals 0, giving roots minus 1 and minus 4, both negative, with sum minus 5 and product 4 exactly as the relations predict.

• (a)α, β > 0 — This is the exact opposite of the truth and it is the option a candidate picks by reading the positivity of a and b as positivity of the roots. The sum of the roots is minus a, not a, and that minus sign is the whole content of the question. Two positive roots would have a positive sum, so they would require a to be negative, contradicting the hypothesis. The direct argument is even shorter: if every coefficient of the quadratic is positive, then substituting any non-negative value of x gives a strictly positive result, so no non-negative number can be a root. It is worth writing Vieta's relations out with their signs each time rather than trusting memory, since for x squared plus px plus q the sum is minus p and the product is plus q, and the asymmetry between those two signs is the single most common source of error in this topic.
• (b)If α > 0, then β < 0 — This option describes roots that straddle zero, one positive and one negative, and that configuration is exactly what the hypotheses rule out. The product of the roots is b, which is given as positive, so the roots must share a sign; a pair consisting of one positive and one negative number would have a negative product and would require b to be negative. The situation the option contemplates therefore never arises under the stated conditions. It is a useful option to have thought about, because the condition for a quadratic to have roots of opposite signs is a standard result in its own right: for a monic quadratic it is precisely that the constant term is negative, since the constant term is the product of the roots. Here it is positive, and the roots are on the same side of zero.
• (d)−1 < α/β < 1 — The ratio of the two roots is positive, because both roots are negative and a negative divided by a negative is positive, so the lower bound in this option is doing no work at all. The upper bound is the substantive claim, and it is false. The two roots are not labelled in any particular order, so nothing prevents α from being the larger of the two in magnitude, in which case the ratio exceeds 1: with a equal to 5 and b equal to 4 the roots are minus 1 and minus 4, and taking α as minus 4 and β as minus 1 gives a ratio of 4. The claim also fails at the boundary case of equal roots, which the hypotheses permit: with a equal to 2 and b equal to 1 both roots are minus 1, and the ratio is exactly 1, which the strict inequality excludes. A single counterexample disposes of a universally quantified statement, and constructing one from small integers is usually faster than trying to prove the statement true.

For a quadratic equation ax squared plus bx plus c equals 0, Vieta's relations give the sum of the roots as minus b over a and the product as c over a; for the monic form x squared plus px plus q equals 0 they reduce to sum equals minus p and product equals q. Those two relations let the signs of the roots be read off the coefficients without solving anything. A positive product means the roots share a sign; a negative product means they straddle zero; a positive sum with a positive product means both are positive; a negative sum with a positive product means both are negative. The discriminant, b squared minus 4ac, decides whether the roots are real and distinct, real and equal, or complex, and in the monic case with positive coefficients the condition for real roots is that p squared is at least 4q. A second and independent way of reasoning about signs comes from evaluating the polynomial: if every coefficient is positive, then the value at any non-negative argument is positive, so no root can be non-negative — the simplest case of Descartes' rule of signs, which relates the number of positive roots to the number of sign changes in the coefficient sequence. Between them, Vieta's relations and the sign rule answer almost every question about where the roots of a quadratic lie without any need to apply the quadratic formula.

This item is typical of the algebra the paper sets: no computation at all, and a result that follows in two lines from a relation the candidate is expected to have at their fingertips. The design of the option set is worth studying because it is the standard one for questions of this kind. One option is the exact negation of the answer, catching a sign error. One describes a configuration the hypotheses exclude, catching a candidate who has not connected the sign of the constant term to the signs of the roots. One is a plausible-looking inequality that is true in some instances and false in others, catching a candidate who tests a single example and generalises. That last construction is the one to be most alert to, because verifying an option on one example proves nothing; a universally quantified claim needs a proof, while its negation needs only one counterexample. The efficient examination-hall procedure on such items is to establish the true state of affairs first, from the hypotheses, and only then look at the options, rather than to test the options one by one against a worked example. Note that the booklet prints the ratio in the last option as a stacked fraction, reproduced here inline with a solidus.

• For the monic quadratic x squared plus ax plus b equals 0, Vieta's relations give the sum of the roots as minus a and the product as b, so positive a and positive b force a negative sum and a positive product.
• A positive product means the roots share a sign and a negative sum then means both are negative, which is why the roots of x squared plus ax plus b with a and b positive are both negative.
• If every coefficient of a polynomial is positive, its value at any non-negative argument is positive, so it can have no non-negative root; this is the simplest consequence of Descartes' rule of signs.
• Real roots require the discriminant to be non-negative, which for this equation means a squared is at least 4b; equality gives a repeated root, so the two roots may coincide.
• The condition for a monic quadratic to have roots of opposite signs is that its constant term is negative, since the constant term is the product of the roots.
• A universally quantified claim about the roots is disproved by a single counterexample: with a equal to 5 and b equal to 4 the roots are minus 1 and minus 4, and with a equal to 2 and b equal to 1 they are both minus 1.

• Reading positive coefficients as positive roots; the sum of the roots of x squared plus ax plus b is minus a, and the minus sign is what the question is about
• Forgetting that a positive constant term forces the roots to share a sign, so that no configuration with one positive and one negative root is available
• Assuming the roots are labelled in order of size; nothing in the question makes α the smaller in magnitude, so a claim about the ratio α over β must hold for either labelling
• Overlooking the case of equal roots, which the hypothesis of real roots permits and which makes the ratio of the roots exactly 1
• Testing an option on a single example and concluding that it is true; one example can refute a universal claim but cannot establish it

Algebra in this paper is set as short reasoning items rather than as computation: the nature or signs of the roots of a quadratic, a condition on coefficients for some property to hold, the value of a symmetric function of the roots, or an inequality that has to be tested. The tools required are few — the discriminant, Vieta's relations and the ability to evaluate a polynomial at a convenient point — and they are asked repeatedly in different dress. The technique that pays is to derive the true position from the hypotheses before reading the options, since the options are constructed to reward each of the standard errors, and then to look for the option that matches. Where an option is an inequality that is sometimes true, reach for small integer counterexamples rather than attempting a proof; the numbers in these questions are always chosen so that a counterexample with single-digit coefficients exists.

No directly related past PYQ was found.

• practice — not a real PYQ

  If the quadratic equation x² + px + q = 0 has roots of opposite signs, then it must be the case that

  • (a)p > 0
  • (b)q > 0
  • (c)q < 0
  • (d)p < 0

  Answer(c) q < 0 — the product of the roots of a monic quadratic is the constant term, so roots of opposite signs, whose product is negative, require q to be negative. Nothing follows about p, since the sum of the roots may be positive, negative or zero depending on which of the two roots is larger in magnitude; and a negative constant term also guarantees a positive discriminant, so the roots are automatically real.

• practice — not a real PYQ

  For the equation x² + ax + b = 0 with a > 0 and b > 0 to have real roots, it is necessary that

  • (a)a² ≥ 4b
  • (b)a² ≤ 4b
  • (c)a = b
  • (d)b ≥ 4a²

  Answer(a) a² ≥ 4b — the roots are real when the discriminant, a squared minus 4b, is non-negative, and equality gives a repeated root. Combined with the sign analysis, this means that whenever such an equation has real roots both of them are negative, and that they coincide precisely at the boundary case where a squared equals 4b, as with a equal to 2 and b equal to 1, whose repeated root is minus 1.