Always Positive or Always Negative Quadratics

No crossing, then the sign of the leading term.

Positive for every x?

Work out x² + x + 3 for a few values of x. At x = 0 it is 3, at x = −1 it is 1 − 1 + 3 = 3, at x = 1 it is 5, and at x = −2 it is 4 − 2 + 3 = 5. Every value is positive. But trying values can never show that every x gives a positive value, because there are infinitely many values of x to try.

The graph suggests the answer. The curve y = x² + x + 3 never comes down to the x-axis, so its value is positive whatever x is. A proof needs a reason, not a picture, and the discriminant gives one.

xy

The curve y = x² + x + 3 stays above the x-axis for every x.

No crossing: the discriminant

For x² + x + 3, a = 1, b = 1 and c = 3, so the discriminant is b² − 4ac = 1 − 12 = −11. It is below zero, so x² + x + 3 = 0 has no real roots, and the curve never meets the x-axis.

A curve that never meets the x-axis cannot pass from above it to below it, because it would have to cross the axis on the way. So x² + x + 3 keeps one sign for every x. At x = 0 its value is 3, which is positive, so it is positive for every x.

Which sign: the leading coefficient

Now take −x² + x − 3. Here a = −1, b = 1 and c = −3, so the discriminant is 1 − 4 × (−1) × (−3) = 1 − 12 = −11. This curve does not meet the x-axis either, so it also keeps one sign. But with a = −1 the curve opens downward, and it stays below the axis: −x² + x − 3 is always negative.

The sign of a decides which side the curve is on. At x = 10, x² + x + 3 = 100 + 10 + 3 = 113, which is positive, while −x² + x − 3 = −100 + 10 − 3 = −93, which is negative. The further out x goes, the more the x² term outweighs the rest. Since the sign never changes, the sign far out along the axis is the sign everywhere.

So two facts settle it. If the discriminant is below zero, the quadratic keeps one sign. Then a > 0 means it is always positive, and a < 0 means it is always negative.

xy

The curve y = −x² + x − 3 opens downward and stays below the x-axis for every x.

When the curve does cross

For x² − x − 2, the discriminant is (−1)² − 4 × 1 × (−2) = 1 + 8 = 9, which is above zero. The curve crosses the x-axis twice, at x = −1 and x = 2, since x² − x − 2 = (x + 1)(x − 2). Between the crossings the value is negative, as at x = 0, where it is −2. Outside them it is positive. The sign changes at each crossing.

A discriminant of exactly 0 is a case of its own. x² − 2x + 1 = (x − 1)² is never negative, but it is 0 at x = 1, so it is not always positive. It is only never negative.

-12

The curve y = x² − x − 2 crosses the x-axis at −1 and 2. It is below the axis between them and above it outside them.

Choosing a coefficient

For which k is x² + 4x + k always positive? Here a = 1, which is positive, so the quadratic is always positive exactly when its discriminant is below zero: 4² − 4 × 1 × k < 0, which is 16 − 4k < 0.

Add 4k to both sides: 16 < 4k. Divide both sides by 4: 4 < k. So x² + 4x + k is always positive when k > 4. Check k = 5: the discriminant is 16 − 20 = −4, which is below zero. At k = 4 it is exactly 0, and x² + 4x + 4 = (x + 2)² is 0 at x = −2, so k = 4 itself does not work.

The usual mistakes

Trying a few values of x and stopping. Every value you try can be positive while some value you did not try is negative. The discriminant covers every x at once.

Looking only at the discriminant. A negative discriminant says the curve never crosses the axis, not which side it is on. The sign of a says which side.

Calling a quadratic with discriminant 0 always positive. It touches 0 at its repeated root.

A square plus a positive number

There is a second way to show that a quadratic is always positive: write it as a square plus a number. Take x² + 4x + 7. Half of 4, the coefficient of x, is 2, and (x + 2)² = x² + 4x + 4. So x² + 4x + 7 = (x + 2)² + 3.

Check it by expanding: (x + 2)² + 3 = x² + 4x + 4 + 3 = x² + 4x + 7. The square (x + 2)² is never negative, so (x + 2)² + 3 is never less than 3. It is always positive, and its least value is 3, at x = −2. The discriminant agrees: 4² − 4 × 1 × 7 = 16 − 28 = −12, which is below zero, and a = 1 is positive.

Worked example: A Cost of Making Each Mug That Can Never Be Zero

Question A workshop makes mugs. In a week when it makes x hundred mugs, the cost of making each mug is C dollars, where C = x2 − 12x + 50. (a) Show that the cost of making a mug can never be zero. (b) Find the least cost of making a mug, and the number of mugs made in a week that gives it.

  1. 1.Half of 12 is 6, and (x − 6)2 = x2 − 12x + 36, so x2 − 12x = (x − 6)2 − 36.

    0102030405060024681012hundreds of mugs in a week, xcost of a mug in dollars, CC = x2− 12x + 50x2− 12x = (x − 6)2− 36
    0102030405060024681012hundreds of mugs in a week, xcost of a mug in dollars, CC = x2− 12x + 50x2− 12x = (x − 6)2− 36
    Half of 12 is 6, and (x − 6)2 = x2 − 12x + 36, so x2 − 12x = (x − 6)2 − 36.

  2. 2.Substitute this into the formula: C = (x − 6)2 − 36 + 50 = (x − 6)2 + 14.

    0102030405060024681012hundreds of mugs in a week, xcost of a mug in dollars, CC = (x − 6)2− 36 + 50C = (x − 6)2+ 14
    0102030405060024681012hundreds of mugs in a week, xcost of a mug in dollars, CC = (x − 6)2− 36 + 50C = (x − 6)2+ 14
    Substitute it into the formula: C = (x − 6)2 + 14.

  3. 3.The square (x − 6)2 is never negative, so C is at least 14 for every value of x. (a) The cost is always positive, so it can never be zero. The discriminant agrees: b2 − 4ac = 144 − 200 = −56 is negative, so x2 − 12x + 50 = 0 has no real roots.

    0102030405060024681012hundreds of mugs in a week, xcost of a mug in dollars, CC is never below 14(x − 6)2is never negative, so C is at least 14b2− 4ac = 144 − 200 = −56: no real roots
    0102030405060024681012hundreds of mugs in a week, xcost of a mug in dollars, CC is never below 14(x − 6)2is never negative, so C is at least 14b2− 4ac = 144 − 200 = −56: no real roots
    (a) A square is never negative, so C is at least 14 and can never be zero. The graph never meets the x-axis.

  4. 4.(b) C is least when (x − 6)2 = 0, which is when x = 6. The least cost is $14 for each mug, when 600 mugs are made in a week. Check: at x = 5 and at x = 7 the cost is 1 + 14 = $15.

    0102030405060024681012hundreds of mugs in a week, xcost of a mug in dollars, CC is never below 14(6, 14)C is least when x = 6: 600 mugs in a weekleast cost: $14 for each mug
    0102030405060024681012hundreds of mugs in a week, xcost of a mug in dollars, CC is never below 14(6, 14)C is least when x = 6: 600 mugs in a weekleast cost: $14 for each mug
    (b) The least cost is $14 for each mug, at x = 6, which is 600 mugs in a week. At x = 5 and at x = 7 the cost is $15.

Answer: (a) C = (x − 6)2 + 14, which is at least 14, so it is never zero; (b) $14, when 600 mugs are made in a week

Common mistakes

  • Substituting a few values of x, finding that each cost is positive, and stopping there. A few values show nothing about all the others. The completed square (x − 6)2 + 14 is at least 14 for every x.
  • Giving the least cost as $6. The 6 in the bracket is the value of x at which the cost is least, which means 600 mugs. The least cost is the number added outside the bracket, $14.

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