QUADRATIC INEQUALITIES - FORMULAS FOR SOLVING, SOLVED EXAMPLES, EXERCISES
Please study
Linear inequalities before Quadratic Inequalities,
if you have not already done so.
Quadratic Polynomial or Quadratic Expression :
ax2 + bx + c where a ( ≠ 0 ), b, c are constants
which can take real number values,
is called a Quadratic Polynomial or Quadratic Expression in x.
If the Quadratic polynomial or Expression is equated to zero,
it is called a Quadratic Equation.
The two methods to solve Quadratic Equations are by
(i) Factorization of Quadratic Polynomials
to get two linear factors, discussed in
Algebra Factoring.
and by (ii) Formula known as
Quadratic Formula.
We make use of that Formula here.
We also need the knowledge of
Factorization of Quadratic Expressions, here.
Here, we solve the Quadratic Expressions
with the signs
'greater than' (>) or 'less than' (<), or
'greater than or equal to' (≥) or 'less than or equal to' (≤).
These are called Quadratic Inequalities.
Solving Quadratic Inequalities :
Formula 1 of Quadratic Inequalities
(x - α)(x - β) > 0 ⇒ x ∉ (α, β) or x ∈ (-∞,α) ∪ (β, ∞) or α > x > β i.e. x doenot lie between α and β
Proof:
We have α < β
The product of (x - α) and (x - β) is positive only when
both are positive or both are negative.
We can see, If x > β, both are positive.
similarly, If x < α, both are negative.
∴ (x - α)(x - β) > 0 ⇒ x doenot lie between α and β (proved.)
Formula 2 of Quadratic Inequalities
(x - α)(x - β) < 0 ⇒ x ∈ (α, β) or α < x < β i.e. x lies between α and β
Proof:
The product of (x - α) and (x - β) is negative only when
one is positive and the other is negative.
This is possible only when x lies between α and β
∴ (x - α)(x - β) < 0 ⇒ x lies between α and β (proved.)
Formula 3 of Quadratic Inequalities
(x - α)2 > 0 ⇒ x ∈ R - {α} or x ∈ (-∞, α) ∪ (α, ∞) i.e. x takes any real value except α
where R is real number set.
Proof:
We know square of any real number except 0, is positive.
∴ (x - α)2 is always positive
except for x - α = 0 i.e. x = α
Hence the result.
Formula 4 of Quadratic Inequalities
(x - α)2 < 0 ⇒ x ∉ R (the real number set) or x ∉ (-∞, ∞) i.e. x can not any take any real value.
Proof :
We know square of any real number can never be negative.
∴ (x - α)2 < 0 has no solution in the real number set. (Proved.)
Formula 5 of Quadratic Inequalities
If x2 + (b⁄a)x + (c⁄a) = 0, has no real roots ( i.e. has imaginary roots),
then x2 + (b⁄a)x + (c⁄a) > 0 ⇒ x ∈ R (the real number set) or x ∈ (-∞, ∞) i.e. x can take any real value.
Proof :
x2 + (b⁄a)x + (c⁄a) = x2 + 2(b⁄2a)x + (b⁄2a)2 - (b⁄2a)2 + (c⁄a)
= (x + b⁄2a)2 + (c⁄a) - (b2⁄4a2)
= (x + b⁄2a)2 + (4ac - b2)⁄4a2
= (x + b⁄2a)2 - Δ⁄(2a)2 where Δ = b2 - 4ac
But Δ is negative since the quadratic equation has imaginary roots.
and all squares are positve.
∴ x2 + (b⁄a)x + (c⁄a) = (+ ve value) - (-ve value)⁄(+ve value) = +ve for all real values of x.
∴ the solution of x2 + (b⁄a)x + (c⁄a) > 0 is R. (proved.)
Formula 6 of Quadratic Inequalities
If x2 + (b⁄a)x + (c⁄a) = 0, has no real roots ( i.e. has imaginary roots),
then x2 + (b⁄a)x + (c⁄a) < 0 ⇒ x ∉ R (the real number set) or x ∉ (-∞, ∞) i.e. x can not any take any real value.
Proof :
Based on the proof of previous Formula, x2 + (b⁄a)x + (c⁄a) can not be negative
for any real value of x.
Hence x2 + (b⁄a)x + (c⁄a) < 0 has no solution in R. (Proved.)
Method of solving a Quadratic Inequation such as, for a ≠ 0,
ax2 + bx + c > 0 or ax2 + bx + c < 0
STEP 1 :
Divide both sides of the inequation by a.
Then the L.H.S. of the inequation becomes x2 + (b⁄a)x + (c⁄a)
The inequality sign may remain the same (if a is positive),
or may get reversed (if a is negative).
The R.H.S.of the inequation will remain 0.
STEP 2 :
Consider the roots of the quadratic equation x2 + (b⁄a)x + (c⁄a) = 0.
Case (i) : Let the roots be real and unequal.( i.e. the Discriminant, Δ > 0). Let the roots be α and β, such that α < β,
Then, express x2 + (b⁄a)x + (c⁄a) as (x - α)(x - β)
Case (ii) : Let the roots be real and equal. ( i.e. the Discriminant, Δ = 0). Let the equal roots be α
Then, express x2 + (b⁄a)x + (c⁄a) as (x - α)2
Case (iii) : Let the roots be imaginary.( i.e. the Discriminant, Δ < 0).
For the meaning of Discriminant, Δ and
the nature of the roots of a quadratic equation,
you may refer to
Quadratic Formula.
STEP 3 :
Make use of formulas 1 and 2 for case (i) to solve the inequation.
Make use of formulas 3 and 4 for case (ii) to solve the inequation.
Make use of formulas 5 and 6 for case (iii) to solve the inequation.
NOTE : To solve ax2 + bx + c ≥ 0 or ax2 + bx + c ≤ 0,
Make the slight modification to the solution,
considering the equality symbol.
The three steps and this slight modification for the equality symbol,
will be
clear by the following set of solved examples.
Set of Solved Examples :
Quadratic Inequalities
Solved Example 1 : Quadratic Inequalities
Solve the quadratic inequalities
(i) 2 - 5x - 18x2 > 0
(ii) 2 - 5x - 18x2 ≥ 0
(i)Solution:
STEP 1 :
The given inequation is 2 - 5x - 18x2 > 0
⇒ -18x2 - 5x + 2 > 0
Dividing both sides by -18, we get
x2 + (5⁄18)x - 2⁄18 < 0
[As we divided with negative number, '>' became '<'.
See Property 3 of Linear Inequalities.]
STEP 2 :
To find the roots of x2 + (5⁄18)x - 2⁄18 = 0:
Comparing the L.H.S. with ax2 + bx + c, we get
a = 1, b = (5⁄18) and c = -2⁄18
Discriminant = Δ = b2 - 4ac
= (5⁄18)2 - 4(1)(-2⁄18) = {25 + 8(18)}⁄{(18)(18)}
= (169)⁄(324) > 0 ⇒ the roots are real and distinct.
√Δ = √{(169)⁄(324)} = 13⁄18
By quadratic formula, the roots of the equation are given by
x = (-b ± √Δ)⁄(2a)
= {-(5⁄18) ± (13⁄18)}⁄{2(1)} = {(-5+13)⁄18}⁄2 or {(-5-13)⁄18}⁄2
= (8⁄18)⁄2 or (-18⁄18)⁄2 = 2⁄9 or -1⁄2
∴
x2 + (5⁄18)x - 2⁄18 = {x - (2⁄9)}{x - (-1⁄2)}
STEP 3 :
By Formula 2 above,
{x - (2⁄9)}{x - (-1⁄2)} < 0
⇒ x lies between (-1⁄2) and (2⁄9) or x ∈ (-1⁄2, 2⁄9) or -1⁄2 < x < 2⁄9
Thus, the solution of the given inequation 2 - 5x - 18x2 > 0 is
x ∈ (-1⁄2, 2⁄9) or -1⁄2 < x < 2⁄9. Ans.
(ii) To solve 2 - 5x - 18x2 ≥ 0.
The solution of (ii) is slightly different from that of (i).
Put '≤' in place of '<' in the above solution.
Thus, the solution of 2 - 5x - 18x2 ≥ 0 becomes
-1⁄2 ≤ x ≤ 2⁄9 or x ∈ [-1⁄2, 2⁄9]
When we put square brackets in place of circular brackets,
the extreme values are included in the range.
Solved Example 2 : Quadratic Inequalities
Solve the quadratic inequalities
(i) 2 - 5x - 18x2 < 0
(ii) 2 - 5x - 18x2 ≤ 0
(i)Solution:
The same quadratic polynomial as in solved example (i)
is taken with
inequality sign reversed.
STEP 1 :
The given inequation is 2 - 5x - 18x2 < 0
⇒ -18x2 - 5x + 2 < 0
Dividing both sides by -18, we get
x2 + (5⁄18)x - 2⁄18 > 0
[As we divided with negative number, '<' became '>'.
See Property 3 of Algebra Inequalities.]
STEP 2 :
To find the roots of x2 + (5⁄18)x - 2⁄18 = 0:
From Solved Example 1 above,
x = 2⁄9 or -1⁄2
∴
x2 + (5⁄18)x - 2⁄18 = {x - (2⁄9)}{x - (-1⁄2)}
STEP 3 :
By Formula 1 above, {x - (2⁄9)}{x - (-1⁄2)} > 0
⇒ x does not lie between (-1⁄2) and (2⁄9) or
x ∉ (-1⁄2, 2⁄9) or x ∈ (-∞,-1⁄2) ∪ (2⁄9, ∞) or -1⁄2 > x > 2⁄9
Thus, the solution of the given inequation 2 - 5x - 18x2 < 0 is
x ∉ (-1⁄2, 2⁄9) or x ∈ (-∞,-1⁄2) ∪ (2⁄9, ∞) or -1⁄2 > x > 2⁄9. Ans.
(ii) To solve 2 - 5x - 18x2 ≤ 0
The solution of (ii) is slightly different from that of (i).
Put '≥' in place of '>' in the above solution.
The solution of 2 - 5x - 18x2 ≤ 0 becomes
-1⁄2 ≥ x ≥ 2⁄9 or x ∈ (-∞,-1⁄2] ∪ [2⁄9, ∞)
When we put square brackets in place of circular brackets,
the extreme values are included in the range.
Solved Example 3 : Quadratic Inequalities
Solve the quadratic inequalities
(i) 4x - 1 - 4x2 > 0
(ii) 4x - 1 - 4x2 ≥ 0
(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 '<'.
See Property 3 of Algebra Inequalities.]
STEP 2 :
To find the roots of x2 - x + 1⁄4 = 0 :
Comparing the L.H.S. with ax2 + bx + c, we get
a = 1, b = -1 and c = 1⁄4
Discriminant = Δ = b2 - 4ac = (-1)2 - 4(1)(1⁄4) = {1 - 1} = 0
⇒ the roots are real and equal.
By quadratic formula, the roots of the equation are given by
x = (-b ± √Δ)⁄(2a) = {-(-1) ± 0}⁄{2(1)} = 1⁄2 or 1⁄2
∴ x2 - x + 1⁄4 = {x - (1⁄2)}2
STEP 3 :
By Formula 4 above,
{x - (1⁄2)}2 < 0
⇒ x ∉ R (the real number set) or x ∉ (-∞, ∞)
i.e. x can not any take any real value.
∴ x2 - x + 1⁄4 < 0 has no real solution for x.
Thus, 4x - 1 - 4x2 > 0
has no solution in the set of Real numbers. Ans.
(ii) To solve 4x - 1 - 4x2 ≥ 0
We have seen 4x - 1 - 4x2 = 0 has the solution x = 1⁄2
and 4x - 1 - 4x2 > 0 has no solution.
∴ the solution of 4x - 1 - 4x2 ≥ 0 is {1⁄2}.Ans.
NOTE: the brackets { } are used for set of individual elements.
∴ {1⁄2} represents single element 1⁄2.
Solved Example 4 : Quadratic Inequalities
Solve the quadratic inequalities
(i) 4x - 1 - 4x2 < 0
(ii) 4x - 1 - 4x2 ≤ 0
(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 has
the 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 = 1⁄2
and
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 5 : Quadratic Inequalities
Solve the quadratic inequalities
(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:
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 + (b⁄a)x + (c⁄a) = 0,
has no real roots ( i.e. has imaginary roots),
then x2 + (b⁄a)x + (c⁄a) < 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.
Solved Example 6 : Quadratic Inequalities
Solve the quadratic inequalities
(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 + (b⁄a)x + (c⁄a) = 0,
has no real roots ( i.e. has imaginary roots),
then x2 + (b⁄a)x + (c⁄a) > 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 has
the 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.
Exercise : Quadratic Inequalities
- Solve the quadratic inequalities
(i) 16x - 15 - 4x2 > 0
(ii) 16x - 15 - 4x2 ≥ 0
- Solve the quadratic inequalities
(i) 16x - 15 - 4x2 < 0
(ii) 16x - 15 - 4x2 ≤ 0
- Solve the quadratic inequalities
(i) 6x - 1 - 9x2 > 0
(ii) 6x - 1 - 9x2 ≥ 0
- Solve the quadratic inequalities
(i) 6x - 1 - 9x2 < 0
(ii) 6x - 1 - 9x2 ≤ 0
- Solve the quadratic inequalities
(i) x2 + 4x + 5 > 0
(ii) x2 + 4x + 5 ≥ 0
- Solve the quadratic inequalities
(i) x2 + 4x + 5 < 0
(ii) x2 + 4x + 5 ≤ 0
For Answers see at the bottom of the page.
Answers to Exercise : Quadratic Inequalities
(1) (i) 3⁄2 < x < 5⁄2. (ii) 3⁄2 ≤x ≤ 5⁄2
(2) (i) 3⁄2 > x > 5⁄2. (ii) 3⁄2 ≥x ≥ 5⁄2
(3) (i) no solution (ii) 1⁄3 (4) (i) R - {1⁄3} (ii) R
(5) (i) R (ii) R (6) (i) no solution (ii) no solution
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