Why a Quadratic Expression Can Stay Positive for Every Value of x

Without solving $a{x}^2+bx+c=0$, one number — the discriminant — reveals whether the roots are real, equal or complex. The same information decides whether the expression is always positive, where its graph turns, and what its greatest or least value is.

This part covers the nature of roots, the sign of a quadratic expression, graphs with maximum and minimum values, and applications.

**For $a{x}^2+bx+c=0$ with real coefficients, the discriminant $D=b^2-4ac$ gives two distinct real roots when $D>0$, equal real roots when $D=0$, and complex conjugate roots when $D<0$, while the roots always satisfy $\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$.

Nature of roots:**

- $D>0$ — real and distinct; rational if D is a perfect square and a, b, c are rational
- $D=0$ — real and equal, both $-\frac{b}{2a}$
- $D<0$ — complex conjugates

Link with sum and product. $(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta =\frac{D}{a^2}$, so the discriminant measures how far apart the roots are.

Worked example. For $3x^2-5x+2=0$: $D=25-24=1$, a perfect square, so the roots are real, distinct and rational — they are 1 and $\frac{2}{3}$, with sum $\frac{5}{3}$ and product $\frac{2}{3}$.

An everyday example. A ball thrown upward from a terrace reaches a chosen height only if the equation for that height has real roots; a negative discriminant means it never gets that high.

The substance. **Roots of opposite signs need only $\frac{c}{a}<0$** — then D is automatically positive, because $-4ac>0$.

**A quadratic expression $a{x}^2+bx+c$ has the sign of a for every real x when $D<0$, the sign of a except at its single root when $D=0$, and when $D>0$ it has the sign of a outside the roots and the opposite sign between them.

Sign rules:**

- $D<0$ — sign of a for all x
- $D=0$ — sign of a, and zero at $x=-\frac{b}{2a}$
- $D>0$ with roots $\alpha <\beta$ — sign of a for $x<\alpha$ or $x>\beta$; opposite sign for $\alpha <x<\beta$

Worked example. Show that $x^2-4x+7>0$ for all real x. Here $a=1>0$ and $D=16-28=-12<0$, so the expression is always positive.

Worked example 2. For $-x^2+5x-6$, $a=-1$ and the roots are 2 and 3. It is negative for $x<2$ or $x>3$ and positive between them — at $x=2.5$ its value is $-6.25+12.5-6=0.25$.

An everyday example. A stall owner's profit rule shaped like a downward quadratic is positive only between two sales levels — sell too few or too many and the stall loses money.

The substance. The sign of a alone is not enough — $x^2-1$ has $a>0$ yet is negative between $-1$ and $1$.

**The graph of $y=a{x}^2+bx+c$ is a parabola that opens upward when $a>0$ and downward when $a<0$, with its vertex at $x=-\frac{b}{2a}$, where the function takes its least value $-\frac{D}{4a}$ if $a>0$ or its greatest value $-\frac{D}{4a}$ if $a<0$.

Sketching steps:

- Direction of opening from the sign of a
- x-intercepts from the roots — two, one or none according to D

Worked example.** Sketch $y=x^2-6x+5$ and find its minimum.

- $a=1>0$, so it opens upward
- Vertex at $x=\frac{6}{2}=3$, where $y=9-18+5=-4$
- Roots 1 and 5

The minimum value is $-4$ at $x=3$; check with $-\frac{D}{4a}=-\frac{36-20}{4}=-4$.

An everyday example. The path of a lofted cricket shot is a downward parabola; its vertex gives the greatest height and where along the ground it occurs.

The substance. The axis of symmetry lies exactly midway between the roots — here $x=3$ is the average of 1 and 5.

Application problems are solved by writing the quantity as a quadratic function, using the vertex for the greatest or least value and the discriminant to decide whether a required value can be reached at all.

Worked example. A stone thrown upward from the ground has height $h=20t-5t^2$ metres after t seconds.

- Greatest height — vertex at $t=\frac{20}{2\times 5}=2$ s, where $h=40-20=20$ m
- Can it reach 15 m? $5t^2-20t+15=0$ has $D=400-300=100>0$, so yes, at $t=1$ s and $t=3$ s
- Can it reach 25 m? $5t^2-20t+25=0$ has $D=400-500=-100<0$, so no

An everyday example. A farmer fencing a goat pen against a long barn wall with 40 m of wire uses three fenced sides with $2x+y=40$. The area $A=x(40-2x)=40x-2x^2$ is greatest at $x=10$ m, giving $A=200$ square metres.

The substance. The discriminant answers 'is it possible?' and the vertex answers 'what is the best?' — most applications need both.