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Basic Concepts Of Quadratic Expressions and its properties for IIT JEE

Posted on - 18-02-2017

JEE Math QE

IIT JEE

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 > 0.

Let f(x) = ax2 + bx + c where a, b, c, ∈ R (a > 0).

f(x) can be rewritten as f(x) = = a, where
D = b2 - 4ac is 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.

Note that if a > 0, the parabola will be concave upwards and if a < 0 the parabola will be concave downwards 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). This gives rise to the following cases:.

(i) a > 0 and b2- 4ac < 0

Û f(x) > 0 " x ∈ R.

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

(ii) a > 0 and b2 - 4ac = 0

Û f(x) ≥ 0 " x ∈ R.

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

(iii) a > 0 and b2- 4ac > 0.

Let f(x) = 0 has two real roots a and b (a<b).

Then f(x) > 0 " x ∈ (-¥, a)È(b, ¥)

And f(x) < 0 " x ∈ (a, b)

In this case the parabola cuts the x- axis at two point a and b and remains concave.


(iv) a < 0 and b2 - 4ac < 0

Û f(x) < 0 " x ∈ R.

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

(v) a < 0 and b2 – 4ac = 0

Û f(x) £ 0 " x ∈ R.

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

(vi) a < 0 and b2- 4ac > 0

Let f(x)=0 have two real roots a and b (a<b).

Then f(x) < 0 " x ∈ (-¥, a)È(b, ¥)

And f(x) > 0 " x ∈ (a, b)

In this case the parabola cuts the x-axis at two point a and b and remains convex .

Notes:

(i) 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

(ii) 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.

Example 1.

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.

Example 2.

If a, b be roots of ax2 + bx + c = 0 , find the equation whose roots are .

Solution:

Since roots are transfered by the same rule

Let y =
&⇒
x =

Substituting in the given quadratic equation, we get


&⇒
a( 1- 3y)2 + 2by (1 - 3y) + 4cy2 = 0


&⇒
(9a - 6b + 4c) y2 + ( - 6a + 2b) y + a = 0 .

Which is the required quadratic equation.

Example 3

Find the values of x for which < 0.

Solution:

Let f(x) = x2 - 4x + 3 and g(x) = x2 + x + 1

Now in g(x) the coefficient of x2 is positive and D < 0,


&⇒
g(x) is positive for all x.

Hence < 0
&⇒
f(x) < 0


&⇒
x2 - 4x + 3 < 0
&⇒
(x - 1)(x - 3) < 0
&⇒
1 < x < 3.

Example 4

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

Solution:

Let f(x) = ax2 - bx + 5 . .

Since f(x) = 0 does not have 2 distinct real roots, we have either
f(x) ≥ 0 " x ∈ R or f(x) £ 0 " x ∈ R

But f(0) = 5 > 0, so f(x) ≥ 0 " x ∈ R

In particular f(-5) ≥ 0
&⇒
25a +5 b +5 ≥ 0
&⇒
5a +b ≥ - 1

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

 
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