The expression ax2 + bx + c is said to be 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).

f(x) can be rewritten as

$\displaystyle f(x) = a [(x+\frac{b}{2a})^2 + \frac{4ac-b^2}{4a^2}]$

$\displaystyle = a[(x+\frac{b}{2a})^2 -\frac{D^2}{4a^2}]$

The quantity D (D = b2 − 4ac) is known as the discriminant of the quadratic expression

Then y = f(x) represents a parabola

whose axis is parallel to the y-axis , with vertex at

A(−b/2a , −D/4a) .

Note that if a > 0, the parabola will be concave upwards and if a < 0 the parabola will be concave downwards. It depends on the sign of b2 − 4ac that the parabola cuts the x-axis at two points if b2 − 4ac > 0, touches the x-axis if b2 − 4ac = 0 or never intersects with the x-axis if b2 – 4ac < 0.

This gives rise to the following cases:

(i) a > 0 and b2 − 4ac < 0

⇒  f(x) > 0 ∀ x ∈ R.

In this case the parabola always remains concave upward and above the x – axis.

(ii) a > 0 and b2 − 4ac = 0

⇒  f(x) ≥ 0 ∀ x ∈ R.

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

(iii) 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 this case the parabola cuts the x- axis at two point α and β and remains concave upward.

(iv) a < 0 and b2 − 4ac < 0

⇒  f(x) < 0 ∀ x ∈ R.
In this case the parabola remains concave downward and always below the x-axis.

(v) a < 0 and b2 − 4ac = 0

⇒  f(x) ≤ 0 ∀ x ∈ R.

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

(vi) a < 0 and b2 − 4ac > 0

Let f(x) = 0 have two real roots α and β (α < β).
Then f(x) < 0 ∀ x ∈ (-∞, α) ∪ (β, ∞)

And f(x) > 0 ∀ x ∈ (α , β)

In this case the parabola cuts the x-axis at two point α and β and remains concave downward.

Notes:
⋄ If a > 0, then minima of f(x) occurs at x = -b/2a and if a < 0 , then maxima of f(x) occurs at x = -b/2a
⋄ If f(x) = 0 has two distinct real roots, then a.f(d) < 0 if and only if d lies between the roots and a.f(d) > 0 if and only if d lies outside the roots.

Illustration : If P(x) = ax2 + bx + c, and Q(x) = − ax2 + dx + c, ac ≠ 0, then prove that P(x) Q(x) = 0 has at least two real roots.

Solution:

P(x) Q(x) = 0

If P(x) = 0 has real roots ,

then b2 − 4ac ≥ 0 .

If Q(x) = 0 has real roots ,

then d2 + 4ac ≥ 0 .

Now ac ≠ 0. If ac < 0 , b2 − 4ac ≥ 0 .

Hence P(x) = 0 has real roots .

If ac > 0, then d2 + 4ac ≥ 0 .

Hence Q(x) = 0 has real roots .

Hence at least two roots of P(x) Q(x) = 0 are real .

Illustration  : If 2x3 + ax2 + bx + 4 = 0 ( a and b are positive real numbers ) has 3 real roots, then prove that

a + b ≥ 6(21/3 + 41/3).

Solution:

Let α, β, γ be the roots of 2x3 + ax2 + bx + 4 = 0.

Given that all the coefficients are positive, so all the roots will be negative.

Let α1 = − α, α2 = − β , α3 = − γ

⇒ α1 + α2 + α3 = a/2

α1 α2 + α2 α3 + α3α1 = b/2

α1 α2 α3 = 2

Applying A.M. ≥ G.M. , we have

1 + α2 + α3 ) /3 ≥ (α1 α2 α3)1/3 => a ≥ 6 × 21/3

Also (α1α2 + α2α3 + α1α3 )/3 ≥ (α1 α2 α3)2/3

⇒ b ≥ 6 ×41/3

Therefore a + b ≥ 6(21/3 + 41/3)

Exercise 2:

(i) If x2 − ax + 4 > 0 for all real x , then find a.

(ii) If f(x) is a quadratic expression such that f(x)>0 ∀ x∈R , and if g(x) = f(x) + f'(x) + f”(x) , then prove that

g(x) > 0 ∀ x ∈ R.

(iii) If c is positive and 2ax2 + 3bx + 5c = 0 does not have any real roots , then prove that 2a – 3b + 5c > 0 .