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.)
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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 5 of Quadratic Inequalities is thus proved.
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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 negativefor any real value of x. Hence x2 + (b⁄a)x + (c⁄a) < 0 has no solution in R. (Proved.)
Formula 6 of Quadratic Inequalities is thus proved.
Method of solving Quadratic Inequalities :
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 αThen, 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 sets of solved examples on Quadratic inequalities.
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