Nature of Roots & Boundary Conditions

A quadratic equation is any equation that can be written in the standard form

• $a, b, c$ are the coefficients ($a$ the leading coefficient, $c$ the constant term); the condition $a \neq 0$ is what makes it quadratic rather than linear.
• A root (or solution) is a value of $x$ that makes the equation true. Graphically, the real roots are exactly the x-intercepts of the parabola $y = ax^2 + bx + c$.
• A quadratic has at most two roots. If $\alpha$ and $\beta$ are the roots, the equation factors as $a(x-\alpha)(x-\beta) = 0$.
• Integer-valued at every integer forces integer coefficients: if $f(x)=x^2+mx+n$ is an integer for every integer $x$, then $f(0)=n$ is an integer and $f(1)-f(0)-1=m$ is an integer. Test $x=0$, then $x=1$ — two evaluations settle both statements.

Three methods for $ax^2 + bx + c = 0$:

• Factoring: write it as $a(x-\alpha)(x-\beta)=0$ and read off $x=\alpha,\beta$. Works cleanly when the roots are rational.
• Completing the square: force a perfect square, e.g. $x^2 + bx = \left(x+\tfrac{b}{2}\right)^2 - \tfrac{b^2}{4}$, then take square roots.
• Quadratic formula: the general result of completing the square,

For $ax^2 + bx + c = 0$ with real coefficients, the discriminant is $D = b^2 - 4ac$ (also written $\Delta$):

• $D > 0$: two distinct real roots (parabola crosses the x-axis twice).
• $D = 0$: two equal real roots, $x = -\tfrac{b}{2a}$ (parabola touches the axis).
• $D < 0$: two complex conjugate roots, no real root (parabola misses the axis).

A further refinement when $a,b,c$ are rational: if $D$ is a perfect square the roots are rational, otherwise they are irrational conjugates. A graph lying entirely above the x-axis means $a > 0$ and $D < 0$.

When a question says "the roots are equal," write $D = 0$ and simplify — the answer is usually a progression among the coefficients:

• GP test: $a, b, c$ in GP $\iff b^2 = ac$. (So $ax^2+bx+c=0$ with $a,b,c$ in GP has $D = ac - 4ac = -3ac$ — non-real roots whenever $a,c$ have the same sign, and then the roots are in the ratio $1:\omega$, i.e. $\alpha/\beta=\omega$ or $\omega^2$, because dividing through by $c$ with $x=\sqrt{c/a}\,t$ gives $t^2+t+1=0$.)
• HP test: $a, b, c$ in HP $\iff \dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$. This is the most-tested outcome — many "equal roots" problems collapse to $\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$.
• AP test: $a, b, c$ in AP $\iff 2b = a + c$.

For $ax^2+bx+c=0$ with roots $\alpha,\beta$:

• Difference of roots: $|\alpha - \beta| = \sqrt{(\alpha+\beta)^2 - 4\alpha\beta} = \dfrac{\sqrt{D}}{|a|}$. A condition like "the roots differ by $k$" becomes $(\alpha-\beta)^2 = k^2$.
• Ratio of roots: if the roots are in ratio $\lambda : 1$, then $\dfrac{(\alpha+\beta)^2}{\alpha\beta} = \dfrac{(\lambda+1)^2}{\lambda} = \dfrac{b^2}{ac}$. Two equations with the same root-ratio satisfy $\dfrac{b^2}{ac} = \dfrac{q^2}{pr}$.

For $ax^2 + bx + c = 0$: substituting $x = 1$ gives $a + b + c$. Therefore

For $f(x) = ax^2 + bx + c$ (take $a > 0$) and an interval $(p, q)$:

• A root lies between $p$ and $q$ $\iff f(p)\cdot f(q) < 0$ (opposite signs — the curve crosses the axis once between them).
• Both roots lie in $(p, q)$ $\iff$ all three hold: $D \geq 0$, $\ a f(p) > 0$ and $a f(q) > 0$, and the vertex $p < -\tfrac{b}{2a} < q$.

Three reducible shapes, all solved by a substitution then a validity check:

• Modulus: for $|x - a| + \ldots$, split into the two sign cases of the modulus and solve each branch on its own interval; keep only the roots that lie in the branch's interval.
• Radical: for an equation in $\sqrt{x}$, set $u = \sqrt{x} \ge 0$; reject any $u < 0$.
• Biquadratic: for $ax^4 + bx^2 + c = 0$, set $u = x^2 \ge 0$; each valid $u > 0$ gives $x = \pm\sqrt{u}$.

Counting the number and TYPE (rational/irrational) of the real roots is the usual question.