Quadratic expression:

The expression ax² + bx + c is called a real quadratic expression in x where a, b, c are real and a ≠ 0. Let f(x) = ax² + bx + c  where a, b, c, ∈ R (a ≠ 0). Now f(x) may be written as  f(x) = .........(1), where  D = b² - 4ac is the discriminant of the quadratic expression.  From (1) it is defined that f(x) = ax² + bx + c may show a parabola  whose axis is parallel to the y-axis, and vertex is at .

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

Case I:    If a > 0

            Sub case A:  a>0 and b² - 4ac < 0 ⇔ f(x) > 0 ∀ x ∈ R.

In that case the parabola usually remains open upward and above the x-axis.

Sub case B:             a > 0 and b² - 4ac = 0  ⇔  f(x) ≥ 0 ∀ x  ∈ R.

In that type the parabola contact with the x-axis at one point and stays open upward.

Sub case C: a > 0 and b² - 4ac > 0. Let f(x) = 0 has two real roots α and β (α < β). Then f(x) > 0 ∀ x ∈ (-∞, α)∪(β, ∞)and f(x) < 0 ∀ x ∈ (α, β)

In that case the parabola intersect the x- axis at two points a and b and stays open upward.

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

In that case ax² + bx + c has no largest value and it has least value

Case II:    If a > 0

            Sub case A:  a < 0 and b² - 4ac < 0 ⇔ f(x) < 0 ∀ x ∈ R.

In this type the parabola stays open downward and usually below the x-axis.

Sub case B: a < 0  and  b² - 4ac  = 0 ⇔  f(x) ≤ 0 ∀ x ∈ R.

In that type the parabola contact with the x - axis and stays open downward.

Sub case C:  a < 0 and b² - 4ac > 0.Let f(x) = 0 have two real roots α and β (α < β).Thenf(x) <0 ∀ x ∈ (-∞, α)∪(β, ∞)and  f(x) > 0 ∀  x ∈ (α, β).

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 

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

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

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