There, We presented the Derivation Of Quadratic Formula in a lucid way. An Example in applying the formula is given. Another problem for practice is also given.

That knowledge is a prerequisite here.

Here, we explain (derive) the Formulae to decide the Nature of the roots of a Quadratic Equation

We also present Solved Examples and Exercise problems on application of these Formulae.

Nature of the roots of a Quadratic Equation :

By Quadratic Formula, the roots of ax^{2} + bx + c = 0 are α = {-b + √(b^{2} - 4ac)}⁄2a and β = {-b - √(b^{2} - 4ac)}⁄2a Let (b^{2} - 4ac) be denoted by Δ (called Delta). Then α = (-b + √Δ)⁄2a and β = (-b - √Δ)⁄2a The nature of the roots (α and β) depends on Δ Δ ( = b^{2} - 4ac) is called the DISCRIMINANT of ax^{2} + bx + c = 0. Three cases arise depending on the value of Δ (= b^{2} - 4ac) is zero or positive or negative. (i) If Δ ( = b^{2} - 4ac) = 0, then α = -b⁄2a and β = -b⁄2a i.e. the two roots are real and equal. Thus ax^{2} + bx + c = 0 has real and equal roots, if Δ = 0 (ii) If Δ ( = b^{2} - 4ac) > 0, the roots are real and distinct. (ii) (a) If Δ ( = b^{2} - 4ac) is a perfect square, the roots are rational. (ii) (b) otherwise, the roots are irrational. (iii) If Δ ( = b^{2} - 4ac) < 0, √Δ is not real.It is called an imaginary number. ∴ α, β are imaginary when Δ is negative. When Δ is negative, the roots are imaginary.

The roots of ax^{2} + bx + c = 0 (i) are real and equal if Δ ( = b^{2} - 4ac) = 0 (ii) are real and distinct if Δ ( = b^{2} - 4ac) > 0 (ii) (a) If Δ ( = b^{2} - 4ac) is a perfect square, the roots are rational. (ii) (b) otherwise, the roots are irrational. (iii) are imaginary if Δ ( = b^{2} - 4ac) < 0

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(i) The given equation is 5x^{2} - 2x - 7 = 0 Comparing this equation with ax^{2} + bx + c = 0, we get a = 5, b = -2 and c = -7 Discriminant = Δ = b^{2} - 4ac = (-2)^{2} - 4(5)(-7) = 4 + 140 = 144 = 12^{2} Since the Discriminant is positive and a perfect square, the roots of the given equation are real, distinct and rational. Ans.

Solution of Example 1(ii) of Quadratic Solver :

(ii) The given equation is 9x^{2} + 24x + 16 = 0 Comparing this equation with ax^{2} + bx + c = 0, we get a = 9, b = 24 and c = 16 Discriminant = Δ = b^{2} - 4ac = (24)^{2} - 4(9)(16) = 576 - 576 = 0 Since the Discriminant is zero, the roots of the given equation are real and equal. Ans.

Solution of Example 1(iii) of Quadratic Solver :

(iii) The given equation is x^{2} + 6x - 5 = 0 Comparing this equation with ax^{2} + bx + c = 0, we get a = 1, b = 6 and c = -5 Discriminant = Δ = b^{2} - 4ac = (6)^{2} - 4(1)(-5) = 36 + 20 = 56 Since the Discriminant is positive and is not a perfect square, the roots of the given equation are real, distinct and irrational. Ans.

Solution of Example 1(iv) of Quadratic Solver :

(iv) The given equation is x^{2} - x + 5 = 0 Comparing this equation with ax^{2} + bx + c = 0, 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 given equation are imaginary. Ans.
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Example 2 : Quadratic Solver

For what values of k are the roots of kx^{2} + (k - 1)x + (k -1) = 0 equal ?For those values of k, find the equal roots.

Solution of Example 2 of Quadratic Solver :

The given equation is kx^{2} + (k - 1)x + (k -1) = 0 Comparing this equation with ax^{2} + bx + c = 0, we get a = k, b = (k - 1) and c = (k -1) Discriminant = Δ = b^{2} - 4ac = (k - 1)^{2} - 4k(k -1) = (k -1)(k - 1 - 4k) = (k -1)(- 1 - 3k) For roots to be equal, Discriminant = 0 ⇒ (k -1)(- 1 - 3k) = 0 ⇒ (k -1) = 0 or (- 1 - 3k) = 0 ⇒ k = 1 or -3k = 1 ⇒ k = 1 or k = -1⁄3 ⇒ k = 1, -1⁄3 roots = (-b ± 0)⁄2a = {-(k - 1)± 0}⁄2k = (1 - k)⁄2k For k = 1, the equal roots are 0, 0. For k = -1⁄3, the value of the equal roots is {1 - (-1⁄3)}⁄2(-1⁄3) = -2 Thus, for k = 1, the equal roots are 0, 0 and for k = -1⁄3, the equal roots are -2, -2. Ans.

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Solve the following problems of Quadratic Solver :

Find the nature of the roots of the equations,

x^{2} - 5x + 6 = 0

x^{2} - 16x + 64 = 0

x^{2} + 2x - 1 = 0

x^{2} + 4x + 5 = 0

If α, β are the roots of px^{2} + qx + r = 0, find the quadratic equation whose roots are(α + 1⁄β), (β + 1⁄α)

For Answers See at the bottom of the Page.

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