The Discriminant and the Nature of the Roots

Its sign counts the crossings.

The number under the square root

The quadratic formula is x = (−b ± √(b² − 4ac)) / 2a. The expression under the square root, b² − 4ac, is called the discriminant, and it is often written Δ, the Greek capital letter delta.

The sign of the discriminant decides how many real roots the equation has, before any other working is done. There are three cases: above zero, equal to zero, and below zero.

Above zero: two roots

Take x² − 4 = 0, where a = 1, b = 0 and c = −4. The discriminant is 0² − 4 × 1 × (−4) = 16. Its square root is 4, so x = (0 ± 4) / 2, which gives x = 2 and x = −2.

When the discriminant is above zero, its square root is a positive number. Adding it gives one root and subtracting it gives a different one, so there are two different real roots, and the curve crosses the x-axis at two places.

−3−2−1123−6−4−2246xy√Δ = 4x = −2x = +242 distinct real roots

the roots sit √Δ = 4 apart

Drag the vertex until the discriminant is zero

The curve y = x² − 4, with its lowest point as the handle. Slide it up and the two crossings close in on each other. When the discriminant reaches zero, they meet.

Zero: one repeated root

Take x² − 2x + 1 = 0, where a = 1, b = −2 and c = 1. The discriminant is (−2)² − 4 × 1 × 1 = 4 − 4 = 0. The square root of 0 is 0, so the ± adds and subtracts nothing: x = (2 ± 0) / 2 = 1, both times.

The two roots are the same number. Factoring shows why: x² − 2x + 1 = (x − 1)(x − 1) = (x − 1)², the same bracket twice. The root x = 1 is called a repeated root. The curve comes down, touches the x-axis at x = 1, and turns back up without crossing it.

x = 1

The curve y = x² − 2x + 1 touches the x-axis at x = 1, its one repeated root.

Below zero: no real roots

Take x² + 2 = 0, where a = 1, b = 0 and c = 2. The discriminant is 0² − 4 × 1 × 2 = −8. The square of every real number is 0 or more, so no real number squares to −8. The formula cannot be finished with real numbers, and the equation has no real roots.

The curve shows the same thing. y = x² + 2 is never less than 2, so the curve stays above the x-axis and never meets it.

xy

The curve y = x² + 2 never comes down to the x-axis, so x² + 2 = 0 has no real roots.

Beyond the real numbers

There is a larger number system, the complex numbers, which includes numbers whose squares are negative. In that system x² + 2 = 0 does have two roots. Among the real numbers, a discriminant below zero means there are no solutions.

Deciding before solving

For 2x² + 3x − 5 = 0, the discriminant is 3² − 4 × 2 × (−5) = 9 + 40 = 49, which is above zero, so there are two real roots. For x² + 2x + 5 = 0, it is 2² − 4 × 1 × 5 = 4 − 20 = −16, which is below zero, so there are none.

The discriminant also answers questions with an unknown coefficient. For which k does x² + kx + 9 = 0 have equal roots? Equal roots mean the discriminant is 0, so k² − 4 × 1 × 9 = 0, which is k² = 36, so k = 6 or k = −6. Check k = 6: x² + 6x + 9 = (x + 3)², a repeated root at x = −3.

The usual mistakes

Squaring a negative b as a negative number. When b = −2, b² = (−2)² = 4, not −4.

Losing the sign of c. When c is negative, −4ac is positive: for x² − 4, the discriminant is 0 − 4 × 1 × (−4) = 16, not −16.

Reading a discriminant of 0 as "no roots". Zero means one repeated root, where the curve touches the axis. Only a negative discriminant means no real roots.

A rectangle with a fixed perimeter

A rectangle has a perimeter of 24 m. Its two lengths and two widths add to 24 m, so one length and one width add to 12 m. Let the width be x m. Then the length is (12 − x) m, and the area is x(12 − x) square meters.

Can the area be 40 m²? That needs x(12 − x) = 40. Expand and move every term to one side: x² − 12x + 40 = 0. The discriminant is (−12)² − 4 × 1 × 40 = 144 − 160 = −16, which is below zero, so no real width works, and the area cannot be 40 m².

The largest area comes from writing the area with a square in it: 12x − x² = 36 − (x − 6)². Check it by expanding: (x − 6)² = x² − 12x + 36, so 36 − (x − 6)² = 36 − x² + 12x − 36 = 12x − x². The square (x − 6)² is never negative, so the area is never more than 36 m². It is exactly 36 m² when x = 6, which makes the rectangle a square of side 6 m.

Worked example: Whether a Given Length of Fencing Can Enclose a Given Area

Question A farmer has 40 m of fencing to make a rectangular pen, and he uses all of it. (a) Can the pen have an area of 120 m2? (b) What is the greatest area that the pen can have, and what shape is it then?

  1. 1.The perimeter is 40 m, so the length and the width add up to 20 m. Let the width be x m. Then the length is (20 − x) m, and an area of 120 m2 needs x(20 − x) = 120.

    02040608010012014005101520width in meters, xarea in square meters, AA = x(20 − x)A = 120width x, length 20 − xx(20 − x) = 120
    02040608010012014005101520width in meters, xarea, AA = x(20 − x)A = 120width x, length 20 − xx(20 − x) = 120
    The length and the width add up to 20 m, so an area of 120 m2 needs x(20 − x) = 120.

  2. 2.Expand the bracket: 20x − x2 = 120. Add x2 to both sides and subtract 20x from both sides: x2 − 20x + 120 = 0.

    02040608010012014005101520width in meters, xarea in square meters, AA = x(20 − x)A = 12020x − x2= 120x2− 20x + 120 = 0
    02040608010012014005101520width in meters, xarea, AA = x(20 − x)A = 12020x − x2= 120x2− 20x + 120 = 0
    Expand and bring every term to one side: x2 − 20x + 120 = 0.

  3. 3.Here a = 1, b = −20 and c = 120, so the discriminant is b2 − 4ac = (−20)2 − 4 × 1 × 120 = 400 − 480 = −80. It is negative, so the equation has no real roots. (a) No width gives 120 m2, so the pen cannot have that area.

    02040608010012014005101520width in meters, xarea in square meters, AA = x(20 − x)A = 120 is never reachedb2− 4ac = (−20)2− 4 × 1 × 120 = 400 − 480 = −80negative: no real roots, so no width gives 120 m2
    02040608010012014005101520width in meters, xarea, AA = x(20 − x)A = 120 is never reachedb2− 4ac = (−20)2− 4 × 1 × 120 = 400 − 480 = −80negative: no real roots, so no width gives 120 m2
    (a) The discriminant is 400 − 480 = −80. It is negative, so there are no real roots, and the pen cannot have an area of 120 m2.

  4. 4.The area of the pen is A = 20x − x2 = −(x2 − 20x). Half of 20 is 10, and (x − 10)2 = x2 − 20x + 100, so x2 − 20x = (x − 10)2 − 100 and A = 100 − (x − 10)2.

    02040608010012014005101520width in meters, xarea in square meters, AA = x(20 − x)A = 120 is never reachedA = −(x2− 20x), and x2− 20x = (x − 10)2− 100A = 100 − (x − 10)2
    02040608010012014005101520width in meters, xarea, AA = x(20 − x)A = 120 is never reachedA = −(x2− 20x), and x2− 20x = (x − 10)2− 100A = 100 − (x − 10)2
    Complete the square: A = 20x − x2 = 100 − (x − 10)2.

  5. 5.The square (x − 10)2 is never negative, so A is never more than 100, and A = 100 when x = 10. (b) The greatest area is 100 m2, when the pen is a square of side 10 m. Check: a pen of 9 m by 11 m has an area of 99 m2.

    02040608010012014005101520width in meters, xarea in square meters, AA = x(20 − x)A = 120 is never reached(10, 100)A = 100 − (x − 10)2is greatest when x = 10greatest area: 100 m2, a square of side 10 m
    02040608010012014005101520width in meters, xarea, AA = x(20 − x)A = 120 is never reached(10, 100)A = 100 − (x − 10)2is greatest when x = 10greatest area: 100 m2, a square of side 10 m
    (b) The greatest area is 100 m2, when x = 10 and the pen is a square of side 10 m. A pen of 9 m by 11 m has 99 m2.

Answer: (a) No: the discriminant is −80, so no width gives an area of 120 m2; (b) 100 m2, when the pen is a square of side 10 m

Common mistakes

  • Letting the length be (40 − x) m. The 40 m is the whole perimeter, which is two lengths and two widths, so one length and one width add up to 20 m.
  • Working out b2 as −400 because b = −20. The square of a negative number is positive: (−20)2 = 400, and the discriminant is 400 − 480 = −80.

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