if you have not already done so.

There we presented 6 Formulas of
The method of Solving Quadratic Inequalities
is also presented Step wise.

The First three Formulas are applied in
The First set of Problems.

Here we deal with Problems
involving the next three Formulas.

## Solved Example 1 : Quadratic Functions

(i) 4x - 1 - 4x2 < 0 (ii) 4x - 1 - 4x2 ≤ 0

Solving Example 1 of Quadratic Functions :

(i)Solution:
STEP 1 :
The given inequation is 4x - 1 - 4x2 < 0
⇒ -4x2 + 4x - 1 < 0
Dividing both sides by -4, we get
x2 - x + 1⁄4 > 0 [As we divided with negative number, '<' became '>']

STEP 2 :
To find the roots of x2 - x + 1⁄4 = 0:
From Solved Example 3 above, x = 1⁄2, 1⁄2

x2 - x + 1⁄4 = {x - (1⁄2)}2

STEP 3 :
By Formula 3 above, {x - (1⁄2)}2 > 0
x ∈ R - {α} or x ∈ (-∞, α) ∪ (α, ∞)
i.e. x takes any real value except α where R is real number set.
x2 - x + 5 > 0 has any real value except 1⁄2 for x as solution.

Thus, x - x2 - 5 < 0 hasthe set of Real numbers except 1⁄2
as the solution. Ans.

(ii) To solve 4x - 1 - 4x2 ≤ 0
We have seen 4x - 1 - 4x2 = 0 has the solution x = 1⁄2and
4x - 1 - 4x2 < 0 has the set of Real numbers except 1⁄2 as the solution.

∴ the solution of 4x - 1 - 4x2 ≤ 0 is set of Real numbers. Ans.

## Solved Example 2 : Quadratic Functions

(i) x - x2 - 5 > 0 (ii) x - x2 - 5 ≥ 0

Solving Example 2 of Quadratic Functions :

(i)Solution:
STEP 1 :
The given inequation is x - x2 - 5 > 0
⇒ -x2 + x - 5 > 0
Dividing both sides by -1, we get
x2 - x + 5 < 0 [As we divided with negative number, '>' became '<']

STEP 2 :
To find the roots of x2 - x + 5 = 0:

Comparing the L.H.S. with ax2 + bx + c, we get
a = 1, b = -1 and c = 5
Discriminant = Δ = b2 - 4ac = (-1)2 - 4(1)(5) = 1 - 20 = -19
Since the Discriminant is negative, the roots of the equation are imaginary

STEP 3 :
By Formula 6 above, we have
If x2 + (ba)x + (ca) = 0, has no real roots ( i.e. has imaginary roots),then x2 + (ba)x + (ca) < 0 ⇒ x ∉ R (the real number set) or x ∉ (-∞, ∞) i.e. x can not any take any real value.
x2 - x + 5 < 0 has no real solution for x.

Thus, x - x2 - 5 > 0 has no solution in the set of Real numbers. Ans.

(ii) To solve x - x2 - 5 ≥ 0
We can see that even for x - x2 - 5 ≥ 0,
there is no solution in the set of Real numbers. Ans.

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## Solved Example 3 : Quadratic Functions

(i) x - x2 - 5 < 0 (ii) x - x2 - 5 ≤ 0

(i)Solution:
STEP 1 :
The given inequation is x - x2 - 5 < 0
⇒ -x2 + x - 5 < 0
Dividing both sides by -1, we get
x2 - x + 5 > 0 [As we divided with negative number, '<' became '>']

STEP 2 :
To find the roots of x2 - x + 5 = 0 :
We have seen in solved example 3, above that the Discriminant is negative,
and the roots of the equation are imaginary

STEP 3 :
By Formula 5 above, we have
If x2 + (ba)x + (ca) = 0, has no real roots ( i.e. has imaginary roots),then x2 + (ba)x + (ca) > 0 ⇒ x ∈ R (the real number set)
or x ∈ (-∞, ∞) i.e. x can take any real value.
x2 - x + 5 > 0 has any real value for x as solution.
Thus, x - x2 - 5 < 0 hasthe set of Real numbers as the solution. Ans.

(ii) To solve x - x2 - 5 ≤ 0
We can see that even for x - x2 - 5 ≤ 0,
the solution is the set of Real numbers. Ans.

(i) 6x - 1 - 9x2 < 0 (ii) 6x - 1 - 9x2 ≤ 0
(i) x2 + 4x + 5 > 0 (ii) x2 + 4x + 5 ≥ 0
(i) x2 + 4x + 5 < 0 (ii) x2 + 4x + 5 ≤ 0

For Answers see at the bottom of the page.

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