What the roots are telling you
Tonight's idea
Last sheet ended with \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). The part under the root, \(D = b^2 - 4ac\), is the discriminant, and it decides everything before you compute anything: \(D \gt 0\) means two real roots, \(D = 0\) one repeated root (the vertex sits on the \(x\)-axis), \(D \lt 0\) no real roots. (There are still two roots then, but they are complex; see Multiplying is rotating.)
If \(ax^2 + bx + c\) has roots \(r\) and \(s\), it factorises as \(a(x - r)(x - s)\). Expand:
\[a(x - r)(x - s) = ax^2 - a(r+s)x + a\,rs.\]
This is the same polynomial as \(ax^2 + bx + c\), so the coefficients match: \(b = -a(r+s)\) and \(c = a\,rs\). Rearranged:
Vieta's formulas. \(r + s = -\dfrac{b}{a}\) and \(rs = \dfrac{c}{a}\).
So you know the sum and product of the roots without finding them. That checks answers in two seconds, builds an equation from its roots, and computes things like \(r^2 + s^2 = (r+s)^2 - 2rs\). Also \((r-s)^2 = (r+s)^2 - 4rs = D/a^2\), so the gap between the roots is \(\sqrt{D}/|a|\).
Watch one: \(x^2 - 5x + 3 = 0\), without solving it
\(D = 25 - 12 = 13 \gt 0\), so two real roots \(r, s\). Vieta: \(r + s = 5\), \(rs = 3\). Then \(r^2 + s^2 = 25 - 6 = 19\), \(\;\dfrac1r + \dfrac1s = \dfrac{r+s}{rs} = \dfrac53\), \(\;|r - s| = \sqrt{13}\).
Now solve it for real: \(x = \tfrac{5 \pm \sqrt{13}}{2}\). Sum 5, difference \(\sqrt{13}\), as Vieta said.
Drag \(a\), \(b\), \(c\) and watch the discriminant and roots at mathdetours.com/demos/quadratic.html.
Your turn
- Find the discriminant and state how many real roots each has: \(x^2 + 4x + 4\), \(\;2x^2 - 3x + 5\), \(\;x^2 - x - 1\).
- For which values of \(k\) does \(x^2 + kx + 16 = 0\) have (a) exactly one root, (b) no real roots?
- The roots of \(2x^2 + 7x - 4 = 0\) are \(r\) and \(s\). Without solving, find \(r + s\), \(rs\) and \(r^2 + s^2\).
- The line \(y = mx - 4\) touches the parabola \(y = x^2\) at exactly one point. Find both possible values of \(m\) and the point of contact for each. (Set \(x^2 = mx - 4\), move everything to one side, and make the discriminant zero.)
Stretch: the golden ratio by Vieta
Let \(\varphi\) and \(\psi\) be the two roots of \(x^2 - x - 1 = 0\) (\(\varphi\) is the golden ratio). Explain why \(\varphi^2 = \varphi + 1\), then compute \(\varphi^2 + \psi^2\) and \(\varphi^4 + \psi^4\) using nothing but Vieta.
Answers
- \(16 - 16 = 0\): one repeated root (\(x = -2\)). \(9 - 40 = -31\): no real roots. \(1 + 4 = 5\): two real roots.
- \(D = k^2 - 64\). (a) \(k = \pm 8\). (b) \(-8 \lt k \lt 8\).
- \(r + s = -\tfrac72\), \(rs = -2\), \(r^2 + s^2 = \tfrac{49}{4} + 4 = \tfrac{65}{4}\). (The roots are \(\tfrac12\) and \(-4\).)
- \(x^2 - mx + 4 = 0\) must have \(D = m^2 - 16 = 0\), so \(m = \pm 4\). With \(m = 4\) the repeated root is \(x = 2\), contact \((2, 4)\); with \(m = -4\), contact \((-2, 4)\).
Stretch. Any root satisfies \(x^2 = x + 1\), so \(\varphi^2 = \varphi + 1\). Vieta: \(\varphi + \psi = 1\), \(\varphi\psi = -1\). Then \(\varphi^2 + \psi^2 = 1 - 2(-1) = 3\), and \(\varphi^4 + \psi^4 = (\varphi^2+\psi^2)^2 - 2(\varphi\psi)^2 = 9 - 2 = 7\). (Continue and you get the Lucas numbers 1, 3, 4, 7, 11, …)