7.3 The Discriminant

The roots of \(ax^{2}+bx+c=0\) are \[x=\frac {-b\pm \sqrt {b^{2}-4ac}}{2a},\] and writing \(D=b^{2}-4ac\) this is \(x=\frac {-b\pm \sqrt {D}}{2a}\).

Everything about the nature of the roots is decided by \(D\) alone, before any root is calculated. \(D\) is called the discriminant, because it discriminates between the three possible cases. Only the sign of \(D\) matters, since \(\sqrt {D}\) is the one place the roots can differ from each other.

Theorem 7.8 (The discriminant). For \(ax^{2}+bx+c=0\) with \(a,b,c\) real and \(a\neq 0\):

(i).: if \(D>0\) the roots are real and distinct;
(ii).: if \(D=0\) the roots are real and equal — one repeated root;
(iii).: if \(D<0\) there are no real roots; the two roots are complex conjugates.

Note 7.9. Each case has a picture. The graph of \(y=ax^{2}+bx+c\) is a parabola, and the real roots are exactly where it crosses the \(x\)-axis. So \(D>0\) means the parabola cuts the axis twice, \(D=0\) means it touches at one point without crossing — the vertex sits on the axis — and \(D<0\) means it misses the axis entirely, lying wholly above it if \(a>0\) or wholly below if \(a<0\).

The complex case is not a failure. In case (iii) the equation still has two roots; they simply are not real, and because \(a,b,c\) are real they come as a conjugate pair \(\frac {-b}{2a}\pm \frac {\sqrt {\left |D\right |}}{2a}i\). This is chapter 3 doing its work: the reason \(\mathbb {C}\) was built was so that no quadratic is left without roots.

Example 7.10. For what values of \(p\) will \(3x^{2}+px+3=0\) have real and distinct roots?

Solution. Real and distinct means \(D>0\). With \(a=3\), \(b=p\), \(c=3\), \[p^{2}-4(3)(3)>0 \implies p^{2}-36>0 \implies (p-6)(p+6)>0 .\] A product of two factors is positive when both are positive or both are negative, that is when \(p>6\) or \(p<-6\): \[p\in (-\infty ,-6)\cup (6,\infty ).\] Note that \(p=6\) and \(p=-6\) are excluded — there the roots are equal, not distinct.

Example 7.11. For what values of \(k\) will \(kx^{2}+(k+1)x+k=0\) have real and equal roots?

Solution. Real and equal means \(D=0\). With \(a=k\), \(b=k+1\), \(c=k\), \begin {align*} (k+1)^{2}-4(k)(k) & = 0\\ k^{2}+2k+1-4k^{2} & = 0\\ -3k^{2}+2k+1 & = 0\\ -3k^{2}+3k-k+1 & = 0\\ \implies \quad 3k(k-1)+1(k-1) & = 0\\ \implies \quad (3k+1)(k-1) & = 0, \end {align*}

so \(k=-\frac {1}{3}\) or \(k=1\).

Note 7.12. A condition is hiding in this question. The equation is quadratic only if \(a\neq 0\), which here means \(k\neq 0\) — with \(k=0\) it collapses to the linear equation \(x=0\), which has one root but not a repeated one. Neither answer is \(0\), so nothing is lost, but a question of this shape should always be checked against it. Whenever the coefficient of \(x^{2}\) contains the unknown, ask first whether it can vanish.

Example 7.13. Determine the nature of the roots of \(2x^{2}+2x+5=0\).

Solution. With \(a=2\), \(b=2\), \(c=5\), \[D=(2)^{2}-4(2)(5)=4-40=-36<0 ,\] so the roots are complex conjugates and the equation has no real solution. Its graph is a parabola opening upwards that never reaches the \(x\)-axis.

The roots themselves are \[x=\frac {-2\pm \sqrt {-36}}{4}=\frac {-2\pm 6i}{4}=-\frac {1}{2}\pm \frac {3}{2}i ,\] a conjugate pair, as predicted.

[next] [prev] [prev-tail] [front] [up]

Questions on this section

Stuck on something here? Ask below and it stays attached to this topic.