Quadratic Expressions

QUADRATIC EXPRESSION

The expression ax² + bx + c is said to be a real quadratic expression in x where a, b, c are real and a $\ne$ 0. Let f(x) = ax² + bx + c where a, b, c, $\in$ R (a $\ne$ 0). Now f(x) can be rewritten as f(x) = $a\left\{{\left(x+\frac{b}{2a}\right)}^2+\frac{4ac-b^2}{4a^2}\right\}$= a$\left[{\left(x+\frac{b}{2a}\right)}^2-\frac{D}{4a^2}\right]$……….(1), where D = b² – 4ac is the discriminant of the quadratic expression. From (1) it is clear that f(x) = ax² + bx + c will represent a parabola whose axis is parallel to the y-axis, and vertex is at A$\left(-\frac{b}{2a},\frac{-D}{4a}\right)$.

It is also clear that if a > 0, the parabola will be open upward and if a < 0 the parabola will be open downward and it depends on the sign of b² –4ac that the parabola cuts the x-axis at two points ( b²-4ac > 0), touches the x-axis (b²- 4ac = 0) or never intersects with the x-axis(b²-4ac < 0).

Case I: If a > 0

Sub case A: a > 0 and b² - 4ac < 0 $\iff$ f(x) > 0$\forall$x $\in$ R.

In this case the parabola always remains open upward and above the x-axis.

Sub case B: a > 0 and b² – 4ac = 0 $\iff$ f(x) $\ge$ 0 $\forall$x $\in$ R.

In this case the parabola touches the x-axis at one point and remains open upward.

Sub case C: a > 0 and b² - 4ac > 0. Let f(x) = 0 has two real roots $\alpha$ and $\beta$ ($\alpha$ < $\beta$). Then f(x) > 0$\forall$ x $\in$ (-$\infty$, $\alpha$)$\cup$($\beta$, $\infty$)and f(x) < 0 $\forall$x$\in$ ($\alpha$, $\beta$)

In this case the parabola cuts the x- axis at two points $\alpha$ and $\beta$ and remains open upward.

Greatest and least value of a quadratic expression ax² + bx + c when a>0:

In this case ax² + bx + c has no greatest value and it has least value $\frac{4ac-b^2}{4a}$ at x = – $\frac{b}{2a}$.

Case II: If a > 0

Sub case A: a < 0 and b² - 4ac < 0$\iff$ f(x) < 0 $\forall$ x $\in$ R.

In this case the parabola remains open downward and always below the x-axis.

Sub case B: a < 0 and b² – 4ac = 0$\iff$ f(x) $\le$ 0 $\forall$ x $\in$ R.

In this case the parabola touches the x - axis and remains open downward.

Sub case C: a < 0 and b² - 4ac > 0.Let f(x) = 0 have two real roots $\alpha$ and $\beta$ ($\alpha$ < $\beta$).Then f(x) < 0$\forall$ x $\in$ (-$\infty$, $\alpha$)$\cup$($\beta$, $\infty$) and f(x) > 0$\forall$ x $\in$ ($\alpha$, $\beta$).

Greatest and least value of a quadratic expression ax² + bx + c when a<0:

If a < 0, then ax² + bx + c has no least value and it has greatest value $\frac{4ac-b^2}{4a}$ at x = – $\frac{b}{2a}$.

Illustration 1: If min {x² + (a – b)x + (1 – a – b)} > max (-x² + (a +b)x – (1 + a + b)) Prove that a² + b² < 4

Solution: Given min{x² + (a – b)x + (1 – a – b)} > max{-x² + (a +b)x – (1 + a + b)}

$\Rightarrow$ min $\left\{{\left(x+\frac{a-b}{2}\right)}^2+\left(1-a-b\right)-\frac{{\left(a-b\right)}^2}{4}\right\}>\max \left\{{\left(\frac{a+b}{2}\right)}^2-\left(1+a+b\right)-{\left(x-\frac{a+b}{2}\right)}^2\right\}$

$\Rightarrow$ 1 – a – b – $\frac{{\left(a-b\right)}^2}{4}$ > ${\left(\frac{a+b}{2}\right)}^2$ – (1 + a + b)

$\Rightarrow$ a² + b² < 4

Illustration 2: If ax²- bx + 5 = 0 does not have 2 distinct real roots, then find the minimum value of 5a +b.

Key concept: Since the given equation doesn't have two real distinct roots, hence its roots will either be imaginary or real and equal. Now the parabola of f(x)= ax²- bx + 5 will either be above the x-axis or will always be below the x-axis. But f(0)=5>0, hence graph of f(x) will always be above the x-axis. Hence f(x)$\ge$0 for all x.

Solution: Now f(x) $\ge$ 0 $\forall$x $\in$ R

In particular f(-5) $\ge$ 0 $\Rightarrow$ 25a +5 b +5 $\ge$ 0 $\Rightarrow$ 5a +b $\ge$ - 1

Hence the least value of 5a +b is - 1.