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Master: Polynomials and quadratics

Algebra · step 4 of 5

A question to start

You stand on a cliff top 25 m above the sea and throw a stone straight up at 20 m per second. If we round the pull of Earth's gravity to 10 m per second squared, the stone's height above the sea after \(t\) seconds is

\[h = -5t^2 + 20t + 25.\]

When does the stone hit the water, and how high does it go? The \(t^2\) means no straight line can describe this flight. You need a new kind of equation, the quadratic, and new tools to solve it.

The idea

Polynomials

A polynomial is a sum of terms, each a number times a whole-number power of the variable, such as \(4x^3 - x^2 + 7x - 2\). The highest power is the degree. Degree 1 is linear (\(3x + 1\)), degree 2 is quadratic (\(x^2 - 5x + 6\)), degree 3 is cubic. The standard form of a quadratic is

\[ax^2 + bx + c, \qquad a \ne 0.\]

Expanding

To expand a product of brackets, multiply every term in the first bracket by every term in the second:

\[(x + 3)(x + 5) = x^2 + 5x + 3x + 15 = x^2 + 8x + 15.\]

Three patterns are worth knowing by heart:

\[(a + b)^2 = a^2 + 2ab + b^2, \qquad (a - b)^2 = a^2 - 2ab + b^2, \qquad (a + b)(a - b) = a^2 - b^2.\]

The last one is the difference of squares. A rectangle picture shows the first: a square of side \(a + b\) splits into an \(a \times a\) square, a \(b \times b\) square and two \(a \times b\) rectangles.

Factoring: expanding backwards

Factoring writes a polynomial as a product. Always look first for a common factor: \(6x^2 + 9x = 3x(2x + 3)\). For \(x^2 + bx + c\), look for two numbers that multiply to \(c\) and add to \(b\):

\[x^2 + 8x + 15 = (x + 3)(x + 5) \quad \text{because } 3 \times 5 = 15 \text{ and } 3 + 5 = 8.\]

Solving by factoring: the zero product rule

If two numbers multiply to give 0, at least one of them must be 0. So once a quadratic equation reads \((\text{something})(\text{something}) = 0\), each bracket gives a solution. To solve \(x^2 - 5x + 6 = 0\): factor to \((x - 2)(x - 3) = 0\), so \(x = 2\) or \(x = 3\). The solutions of an equation \(p(x) = 0\) are called its roots.

Completing the square

Not every quadratic factors neatly. Completing the square always works. Use \((x + k)^2 = x^2 + 2kx + k^2\) backwards: take half the coefficient of \(x\), square it, and add and subtract it.

\[x^2 + 6x + 2 = (x^2 + 6x + 9) - 9 + 2 = (x + 3)^2 - 7.\]

A square is never negative, so the smallest value of \((x + 3)^2 - 7\) is \(-7\), reached at \(x = -3\).

The quadratic formula

Completing the square on the general equation \(ax^2 + bx + c = 0\) gives a formula that solves every quadratic. Divide by \(a\), move \(\tfrac{c}{a}\) across, and add \(\left(\tfrac{b}{2a}\right)^2\) to both sides:

\[ \begin{aligned} x^2 + \frac{b}{a}x &= -\frac{c}{a}\\ \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}\\ x + \frac{b}{2a} &= \pm\frac{\sqrt{b^2 - 4ac}}{2a} \end{aligned} \]

and so

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]

The \(\pm\) ("plus or minus") gives two answers, one with \(+\) and one with \(-\).

The discriminant

The part under the square root, \(\Delta = b^2 - 4ac\) (Δ is the Greek capital delta), tells you how many real solutions there are before you do any more work:

  • \(\Delta > 0\): two different real solutions;
  • \(\Delta = 0\): one (repeated) solution;
  • \(\Delta < 0\): no real solutions, because no real number squares to a negative.

Parabolas

The graph of \(y = ax^2 + bx + c\) is a parabola, a symmetric U-shaped curve. It opens upward if \(a > 0\) and downward if \(a < 0\). Its turning point, the vertex, sits on the line of symmetry

\[x = -\frac{b}{2a}.\]

In vertex form \(y = a(x - p)^2 + q\), the vertex is simply \((p, q)\). The roots are where the parabola crosses the \(x\)-axis; the discriminant tells you whether it crosses twice, touches once, or misses.

Worked examples

1. The stone from the cliff. Solve \(-5t^2 + 20t + 25 = 0\).

  • Divide every term by \(-5\): \(t^2 - 4t - 5 = 0\).
  • Two numbers that multiply to \(-5\) and add to \(-4\): \(-5\) and \(1\). So \((t - 5)(t + 1) = 0\).
  • \(t = 5\) or \(t = -1\). Time cannot be negative here, so the stone hits the water after 5 seconds.
  • Highest point: the vertex is at \(t = -\dfrac{20}{2 \times (-5)} = 2\) seconds, where \(h = -5(4) + 40 + 25 = 45\) m above the sea.

On Mars, gravity is about 3.7 m per second squared, so the \(-5\) becomes about \(-1.85\). Solving \(-1.85t^2 + 20t + 25 = 0\) with the formula gives \(t \approx 11.9\) seconds, and the stone rises to about \(25 + \dfrac{20^2}{4 \times 1.85} \approx 79\) m. Same equation shape, a gentler planet.

2. The formula. Solve \(2x^2 + 3x - 2 = 0\).

  • \(a = 2\), \(b = 3\), \(c = -2\). Discriminant: \(\Delta = 9 - 4(2)(-2) = 9 + 16 = 25\).
  • \(x = \dfrac{-3 \pm 5}{4}\), so \(x = \dfrac{2}{4} = \dfrac12\) or \(x = \dfrac{-8}{4} = -2\).
  • Check \(x = \tfrac12\): \(2 \times \tfrac14 + \tfrac32 - 2 = \tfrac12 + \tfrac32 - 2 = 0\).

3. Vertex form. Write \(y = x^2 - 6x + 5\) in vertex form, and find its vertex and roots.

  • Half of \(-6\) is \(-3\); \((-3)^2 = 9\). So \(y = (x^2 - 6x + 9) - 9 + 5 = (x - 3)^2 - 4\).
  • Vertex \((3, -4)\). It opens upward, so \(-4\) is the minimum.
  • Roots: \((x - 3)^2 = 4\) gives \(x - 3 = \pm 2\), so \(x = 1\) or \(x = 5\), both 2 steps either side of the line of symmetry \(x = 3\).

Try it

  1. Expand \((x + 4)(x - 3)\).
  2. Factor \(x^2 - 9x + 20\).
  3. Solve \(x^2 - 49 = 0\).
  4. Solve \(3x^2 - 5x - 1 = 0\), giving answers to 2 decimal places.
  5. How many real solutions does \(x^2 + 4x + 7 = 0\) have?
  6. Find the vertex of \(y = x^2 - 8x + 3\).
Answers
  1. \(x^2 - 3x + 4x - 12 = x^2 + x - 12\).
  2. Two numbers that multiply to 20 and add to \(-9\): \(-4\) and \(-5\). So \((x - 4)(x - 5)\).
  3. Difference of squares: \((x - 7)(x + 7) = 0\), so \(x = 7\) or \(x = -7\).
  4. \(\Delta = 25 - 4(3)(-1) = 37\). \(x = \dfrac{5 \pm \sqrt{37}}{6}\), so \(x \approx 1.85\) or \(x \approx -0.18\).
  5. None. \(\Delta = 16 - 28 = -12 < 0\). (In vertex form it is \((x + 2)^2 + 3\), which is always at least 3.)
  6. Half of \(-8\) is \(-4\), and \((-4)^2 = 16\), so \(y = (x - 4)^2 - 16 + 3 = (x - 4)^2 - 13\). The vertex is \((4, -13)\).

Common mistake

Thinking \((a + b)^2 = a^2 + b^2\). It is not: the middle term \(2ab\) is missing. Test with numbers: \((3 + 4)^2 = 49\), but \(3^2 + 4^2 = 25\). The square of a sum always has the two "rectangles" \(ab\) in it. A close cousin: in the quadratic formula, the whole of \(-b \pm \sqrt{\Delta}\) is divided by \(2a\), not just the square root.

Where it leads