Here we deal with Problems involving the next three Formulas.

Solved Example 1 : Quadratic Functions

Solve the quadratic inequalities (i) 4x - 1 - 4x^{2} < 0 (ii) 4x - 1 - 4x^{2} ≤ 0

Solving Example 1 of Quadratic Functions :

(i)Solution: STEP 1 : The given inequation is 4x - 1 - 4x^{2} < 0 ⇒ -4x^{2} + 4x - 1 < 0 Dividing both sides by -4, we get x^{2} - x + 1⁄4 > 0 [As we divided with negative number, '<' became '>']

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

∴ x^{2} - 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. ∴ x^{2} - x + 5 > 0 has any real value except 1⁄2 for x as solution.

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

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

∴ the solution of 4x - 1 - 4x^{2} ≤ 0 is set of Real numbers. Ans.

Solved Example 2 : Quadratic Functions

Solve the quadratic inequalities (i) x - x^{2} - 5 > 0 (ii) x - x^{2} - 5 ≥ 0

Solving Example 2 of Quadratic Functions :

(i)Solution: STEP 1 : The given inequation is x - x^{2} - 5 > 0 ⇒ -x^{2} + x - 5 > 0 Dividing both sides by -1, we get x^{2} - x + 5 < 0 [As we divided with negative number, '>' became '<']

STEP 2 : To find the roots of x^{2} - x + 5 = 0:

Comparing the L.H.S. with ax^{2} + bx + c, we get a = 1, b = -1 and c = 5 Discriminant = Δ = b^{2} - 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 x^{2} + (b⁄a)x + (c⁄a) = 0, has no real roots ( i.e. has imaginary roots),then x^{2} + (b⁄a)x + (c⁄a) < 0 ⇒ x ∉ R (the real number set) or x ∉ (-∞, ∞) i.e. x can not any take any real value. ∴ x^{2} - x + 5 < 0 has no real solution for x.

Thus, x - x^{2} - 5 > 0 has no solution in the set of Real numbers. Ans.

(ii) To solve x - x^{2} - 5 ≥ 0 We can see that even for x - x^{2} - 5 ≥ 0, there is no solution in the set of Real numbers. Ans.

Solve the quadratic inequalities (i) x - x^{2} - 5 < 0 (ii) x - x^{2} - 5 ≤ 0

(i)Solution: STEP 1 : The given inequation is x - x^{2} - 5 < 0 ⇒ -x^{2} + x - 5 < 0 Dividing both sides by -1, we get x^{2} - x + 5 > 0 [As we divided with negative number, '<' became '>']

STEP 2 : To find the roots of x^{2} - 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 x^{2} + (b⁄a)x + (c⁄a) = 0, has no real roots ( i.e. has imaginary roots),then x^{2} + (b⁄a)x + (c⁄a) > 0 ⇒ x ∈ R (the real number set) or x ∈ (-∞, ∞) i.e. x can take any real value. ∴ x^{2} - x + 5 > 0 has any real value for x as solution. Thus, x - x^{2} - 5 < 0 hasthe set of Real numbers as the solution. Ans.

(ii) To solve x - x^{2} - 5 ≤ 0 We can see that even for x - x^{2} - 5 ≤ 0, the solution is the set of Real numbers. Ans.

Exercise : Quadratic Functions

Solve the quadratic inequalities (i) 6x - 1 - 9x^{2} < 0 (ii) 6x - 1 - 9x^{2} ≤ 0

Solve the quadratic inequalities (i) x^{2} + 4x + 5 > 0 (ii) x^{2} + 4x + 5 ≥ 0

Solve the quadratic inequalities (i) x^{2} + 4x + 5 < 0 (ii) x^{2} + 4x + 5 ≤ 0

For Answers see at the bottom of the page.

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