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:.
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(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.
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|
(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.
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(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.
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|
(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.
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(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.
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|
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(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 .
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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.
