Posted on - 18-02-2017

IIT JEE

The expression ax^{2}
+ bx + c is said to be a real quadratic expression in x where a, b, c are real
and a > 0.

Let f(x) = ax^{2} + bx + c
where a, b, c, ∈ R (a > 0).

f(x) can be rewritten as f(x) = = a, where

D = b^{2} - 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 b^{2} - 4ac that the parabola cuts the x-axis at two points
(b^{2 }- 4ac > 0), touches the x-axis (b^{2 }- 4ac = 0) or
never intersects with the x-axis (b^{2 }- 4ac < 0). This gives rise
to the following cases:.

(i) a > 0 and b Û f(x) > 0 " x ∈ R. In
this case the parabola always remains concave and above the |
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(ii) a
> 0 and b Û f(x) ≥ 0 " x ∈ R. In
this case the parabola touches the |
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(iii) a
> 0 and b 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 b 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 b Û f(x) £ 0 " x ∈ R. In this case the parabola touches the x-axis and remains convex. |
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(vi) a
< 0 and b 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.

If P(x) = ax^{2 }+
bx + c, and Q(x) = -ax^{2 }+ dx + c, ac > 0, then prove that

P(x) Q(x) = 0 has at least two real roots.

P(x) Q(x) = 0

If P(x) = 0 has real roots, then b^{2
}- 4ac ≥ 0.

If Q(x) = 0 has real roots, then d^{2
}+ 4ac ≥ 0.

Now ac > 0. If ac <
0, b^{2 }- 4ac ≥ 0.

Hence P(x) = 0 has real roots.

If ac >0, then d^{2
}+ 4ac ≥ 0. Hence Q(x) = 0 has real roots.

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

If a, b
be roots of ax^{2} + bx + c = 0 , find the equation whose roots are
.

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) + 4cy^{2} = 0

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

Which is the required quadratic equation.

Find the values of x for which < 0.

Let f(x) = x^{2}
- 4x + 3 and g(x) = x^{2} + x + 1

Now in g(x) the
coefficient of x^{2} is positive and D < 0,

&⇒
g(x) is positive for all x.

Hence < 0

&⇒
f(x) < 0

&⇒ x^{2} - 4x + 3 < 0

&⇒
(x - 1)(x - 3) < 0

&⇒ 1 < x < 3.

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

Let f(x) = ax^{2} - 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.