Quadratic Equations & Theory of Equations
The parabola, the discriminant, and the deep link between a polynomial's roots and its coefficients
- The quadratic formula, and how the discriminant alone decides the nature of the roots.
- Vieta's relations — reading the sum and product of roots straight off the coefficients.
- How to read everything off the parabola: sign, vertex, maximum/minimum and range.
- The conditions for a common root and for locating roots relative to a number.
- To solve quadratic inequalities with the wavy-curve method.
- The general theory of equations: roots-and-coefficients for any degree, conjugate roots, and Descartes' rule of signs.
From Numbers to Equations
A polynomial equation sets a polynomial equal to zero, and its solutions are the roots. The humblest interesting case is the quadratic — degree two — and it repays close study, because every technique here scales up to cubics, quartics and beyond.
The Fundamental Theorem of Algebra (from Chapter 2) promises that a degree-\(n\) equation has exactly \(n\) roots, counted with multiplicity, in \(\mathbb{C}\). Our job is to find them, and — even when we cannot find them cleanly — to describe them: how many are real, what their sum and product are, where they sit on the number line.
The Quadratic Equation & Its Roots
A quadratic equation is any equation of the form
The two roots are usually written \(\alpha\) and \(\beta\). The quantity under the root governs everything, so it gets its own name.
The Discriminant: Nature of the Roots
Its sign tells you, without solving, exactly what kind of roots you have — a fact the parabola makes obvious: \(D\) simply counts how many times the curve meets the \(x\)-axis.
| Discriminant | Roots | Refinement (rational a,b,c) |
|---|---|---|
| \(D>0\) | real and distinct | rational if \(D\) is a perfect square, else irrational conjugates \(p\pm\sqrt q\) |
| \(D=0\) | real and equal, \(x=-\dfrac{b}{2a}\) | always rational |
| \(D<0\) | complex conjugates \(p\pm iq\) | — |
Sum & Product of Roots (Vieta)
You rarely need the roots themselves — you need their sum and product, and those are sitting right there in the coefficients. Comparing \(ax^2+bx+c=a(x-\alpha)(x-\beta)\) gives Vieta's relations:
Also useful: \(|\alpha-\beta|=\dfrac{\sqrt{D}}{|a|}\), and any symmetric expression in \(\alpha,\beta\) can be rebuilt from these two — e.g. \(\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta\) and \(\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{\alpha+\beta}{\alpha\beta}\).
Building a Quadratic from Its Roots
Run Vieta in reverse. If you know the sum \(S\) and product \(P\) of the desired roots, the (monic) quadratic is built instantly:
The Graph of a Quadratic
The graph of \(y=ax^{2}+bx+c\) is a parabola. Its direction is set by the sign of \(a\), and its turning point — the vertex — sits on the axis of symmetry \(x=-\dfrac{b}{2a}\).
| Feature | Value / rule |
|---|---|
| Opens | upward if \(a>0\), downward if \(a<0\) |
| Axis of symmetry | \(x=-\dfrac{b}{2a}\) |
| Vertex | \(\left(-\dfrac{b}{2a},\ -\dfrac{D}{4a}\right)\) |
| y-intercept | \((0,\,c)\) |
When \(D>0\), the expression has the sign of \(a\) outside the roots and the opposite sign between them. This single rule drives every inequality in §3-10.
Maximum, Minimum & Range
The vertex is the extreme value. If the parabola opens up, the vertex is the lowest point — a minimum; if it opens down, a maximum.
So the range over all real \(x\) is \(\left[-\dfrac{D}{4a},\,\infty\right)\) when \(a>0\), and \(\left(-\infty,\,-\dfrac{D}{4a}\right]\) when \(a<0\).
Common Roots
When two quadratics share a root, the coefficients must satisfy a relation. Let \(a_1x^{2}+b_1x+c_1=0\) and \(a_2x^{2}+b_2x+c_2=0\).
| Situation | Condition |
|---|---|
| One common root | \((c_1a_2-c_2a_1)^{2}=(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)\) |
| Both roots common | \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\) |
Location of Roots
Sometimes you do not need the roots — only where they lie relative to a number \(k\). Three quantities settle every such question: the discriminant \(D\), the sign of \(a\cdot f(k)\), and the position of the vertex \(-\dfrac{b}{2a}\) relative to \(k\). Here \(f(x)=ax^{2}+bx+c\).
| Requirement on the roots | Conditions (all must hold) |
|---|---|
| Both roots \(> k\) | \(D\ge0,\quad a\,f(k)>0,\quad -\dfrac{b}{2a}>k\) |
| Both roots \(< k\) | \(D\ge0,\quad a\,f(k)>0,\quad -\dfrac{b}{2a}<k\) |
| \(k\) lies between the roots | \(a\,f(k)<0\) (this alone) |
| Both roots in \((k_1,k_2)\) | \(D\ge0,\ a\,f(k_1)>0,\ a\,f(k_2)>0,\ k_1<-\dfrac{b}{2a}<k_2\) |
| Exactly one root in \((k_1,k_2)\) | \(f(k_1)\,f(k_2)<0\) |
\(a\,f(k)>0\) says the point \(k\) is outside the interval of the roots (same side as the arms of the parabola); \(a\,f(k)<0\) says it is inside (in the dip below the axis). Sketch the parabola and the conditions write themselves.
Quadratic Inequalities
To solve \(ax^{2}+bx+c>0\) (or \(<0\)), factor it and use the sign rule from §3-6. The fastest organiser is the wavy-curve (sign) method: mark the roots on a number line and alternate signs between them.
Theory of Equations: Higher Polynomials
Vieta's idea generalises to any degree. For the polynomial equation \(a_0x^{n}+a_1x^{n-1}+\dots+a_{n}=0\) with roots \(\alpha_1,\dots,\alpha_n\):
The elementary symmetric sums alternate in sign: take coefficients in order, divide by the leading one, and flip the sign at every step.
| Symmetric sum | Value |
|---|---|
| \(\alpha+\beta+\gamma\) | \(-\dfrac{b}{a}\) |
| \(\alpha\beta+\beta\gamma+\gamma\alpha\) | \(\dfrac{c}{a}\) |
| \(\alpha\beta\gamma\) | \(-\dfrac{d}{a}\) |
Apply it to \(f(-x)\) to bound the negative real roots. A quick way to predict how many real roots to hunt for.
Symmetric Functions & Transformation of Equations
If an equation has roots \(\alpha,\beta,\gamma,\dots\), you can write down a new equation whose roots are a chosen function of the old ones — without finding the roots. The trick is a substitution.
| New roots wanted | Substitute |
|---|---|
| Negatives \(-\alpha\) | replace \(x\) by \(-x\) |
| Reciprocals \(1/\alpha\) | replace \(x\) by \(1/x\) (reverse the coefficients) |
| Each root \(+h\) | replace \(x\) by \(x-h\) |
| Each root scaled by \(k\) | replace \(x\) by \(x/k\) |
Putting It to Work
Problem. Find \(k\) so that \(x^{2}-4x+k=0\) has equal roots, and give that repeated root.
Solution. Equal roots need \(D=0\): \((-4)^{2}-4k=0\Rightarrow k=4\). The repeated root is \(x=-\dfrac{b}{2a}=\dfrac{4}{2}=2\).
Problem. If \(\alpha,\beta\) are roots of \(2x^{2}-5x+1=0\), find \(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\).
Solution. By Vieta, \(\alpha+\beta=\tfrac{5}{2}\), \(\alpha\beta=\tfrac{1}{2}\). Now
Problem. If \(\alpha,\beta\) are roots of \(x^{2}-6x+8=0\), form the equation whose roots are \(\alpha^{2},\beta^{2}\).
Solution. Here \(\alpha+\beta=6,\ \alpha\beta=8\). New sum \(S=\alpha^{2}+\beta^{2}=36-16=20\); new product \(P=(\alpha\beta)^{2}=64\). So the required equation is
Problem. Find the range of \(y=\dfrac{x^{2}+x+1}{x^{2}+1}\) for real \(x\).
Solution. Clear the denominator: \(y(x^{2}+1)=x^{2}+x+1\Rightarrow(y-1)x^{2}-x+(y-1)=0\). For real \(x\) we need \(D\ge0\):
So the range is \(\left[\tfrac12,\tfrac32\right]\). (The degenerate case \(y=1\) gives \(x=0\), which is inside this interval.)
Problem. For what values of \(m\) does \(x^{2}-mx+(2m-3)=0\) have one root less than 1 and the other greater than 1?
Solution. The number 1 lies between the roots iff \(a\,f(1)<0\). Here \(a=1\) and \(f(1)=1-m+(2m-3)=m-2\). So the condition is \(1\cdot(m-2)<0\), giving \(m<2\). (Note: \(a\,f(1)<0\) automatically forces \(D>0\), so the roots are guaranteed real and distinct.)
Problem. If \(\alpha,\beta,\gamma\) are roots of \(x^{3}-3x^{2}+x-1=0\), find \(\alpha^{2}+\beta^{2}+\gamma^{2}\).
Solution. From Vieta: \(\sum\alpha=3,\ \sum\alpha\beta=1\). Then
Chapter Summary
\(x=\frac{-b\pm\sqrt D}{2a}\); \(D=b^2-4ac\) fixes the nature of the roots. Conjugate (complex or surd) roots come in pairs.
\(\alpha+\beta=-b/a\), \(\alpha\beta=c/a\). Build any symmetric expression — and any new equation \(x^2-Sx+P=0\) — from these.
Direction from \(a\), vertex at \((-b/2a,-D/4a)\). Sign of \(a\) governs sign and extreme value; range follows at once.
Cross-multiplication for a common root; \(D,\ a\,f(k),\ -b/2a\) together pin roots relative to any \(k\).
Factor and apply the wavy-curve sign rule: positive outside the roots, negative between (for \(a>0\)).
Roots-and-coefficients for any degree (alternating signs), conjugate roots, transformations, and Descartes' rule of signs.
Problems
Reach for Vieta and the discriminant before solving; let the parabola guide every sign question. Difficulty rises down the list.
- Discuss the nature of the roots of \(3x^{2}-2x+1=0\) and of \(2x^{2}-7x+3=0\).
- For what value of \(m\) does \((m+1)x^{2}+2(m+3)x+(m+8)=0\) have equal roots?
- If \(\alpha,\beta\) are roots of \(x^{2}+px+q=0\), find \(\alpha^{3}+\beta^{3}\) in terms of \(p,q\).
- Form the quadratic whose roots are the reciprocals of the roots of \(3x^{2}-5x+2=0\).
- Find the minimum value of \(2x^{2}+3x+5\) and the value of \(x\) where it occurs.
- Find the range of \(y=\dfrac{x^{2}-2x+4}{x^{2}+2x+4}\) for real \(x\).
- If \(x^{2}+ax+b=0\) and \(x^{2}+bx+a=0\) have a common root (with \(a\neq b\)), show that \(a+b+1=0\).
- Solve the inequality \(\dfrac{x-1}{x+2}\ge0\).
- Find all \(k\) for which both roots of \(x^{2}-2kx+(k^{2}-1)=0\) lie between \(-2\) and \(4\).
- Use Descartes' rule of signs to bound the number of positive and negative real roots of \(x^{4}-x^{3}+x^{2}-x+1=0\).
- If \(\alpha,\beta,\gamma\) are roots of \(x^{3}+px+q=0\), find \(\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}\) and \(\alpha^{2}+\beta^{2}+\gamma^{2}\).
- If one root of \(x^{2}-px+q=0\) is the square of the other, prove that \(p^{3}-q(3p+1)-q^{2}=0\).