There, We presented the Derivation Of Formulae Relating roots and coefficients of a Quadratic Equation. Solved Examples on applying the formulae are given. Problems for practice are also given in Exercise.

Here we present some more Solved Examples and Exercise problems.

Example 1 : The Quadratic Formula

Solved Example 1 of The Quadratic Formula :

If α, β are the roots of x^{2} - px + q = 0, then find the value of

α^{3} + β^{3}

(α^{2}⁄β) + (β^{2}⁄α)

(1⁄α^{3}) + (1⁄β^{3})

(α⁄β) + (β⁄α)

Solution of Example 1 of The Quadratic Formula :

The given equation is x^{2} - px + q = 0 Comparing this equation with ax^{2} + bx + c = 0, we get a = 1, b = -p and c = q α, β are the roots of x^{2} - px + q = 0 We know, Sum of the roots = α + β = -b⁄a = -(-p)⁄1 = p Product of the roots = αβ = c⁄a = q ⁄1 = q

⇒ a^{3} + b^{3} = (a + b)^{3} - 3ab(a + b) Applying this here, we get α^{3} + β^{3} = (α + β)^{3} - 3αβ( α + β) Using the values of α + β (= p) and αβ (= q) here, we get α^{3} + β^{3} = ( p)^{3} - 3q(p) = p^{3} - 3pq Ans.

(ii) To find the value of (α^{2}⁄β) + (β^{2}⁄α)

Let A = (α^{2}⁄β) + (β^{2}⁄α) Multiplying both sides with αβ, we get αβA = αβ(α^{2}⁄β) + αβ(β^{2}⁄α)= α^{3} + β^{3} Using the values of α^{3} + β^{3} (= p^{3} - 3pq) and αβ (= q) here, we get qA = p^{3} - 3pq ⇒ A = (p^{3} - 3pq)⁄q⇒ (α^{2}⁄β) + (β^{2}⁄α) = (p^{3} - 3pq)⁄q. Ans.

(iii) To find the value of (1⁄α^{3}) + (1⁄β^{3})

Let A = (1⁄α^{3}) + (1⁄β^{3}) Multiplying both sides with α^{3}β^{3}, we get α^{3}β^{3}A = α^{3}β^{3}(1⁄α^{3}) + α^{3}β^{3} (1⁄β^{3}) (αβ)^{3}A = β^{3} + α^{3} = α^{3} + β^{3} Using the values of α^{3} + β^{3} (= p^{3} - 3pq) and αβ (= q) here, we get q^{3}A = p^{3} - 3pq ⇒ A = (p^{3} - 3pq)⁄q^{3}⇒ (1⁄α^{3}) + (1⁄β^{3}) = (p^{3} - 3pq)⁄q^{3}. Ans.

(iv) To find the value of (α⁄β) + (β⁄α) Let A = (α⁄β) + (β⁄α) Multiplying both sides with αβ, we get αβA = αβ(α⁄β) + αβ(β⁄α)= α^{2} + β^{2} = (α + β)^{2} - 2αβ Using the values of α + β (= p) and αβ (= q) here, we get qA = (p)^{2} - 2q ⇒ A = (p^{2} - 2q)⁄q ⇒ (α⁄β) + (β⁄α) = (p^{2} - 2q)⁄q. Ans.

Thus all the four problems of The Quadratic Formula are solved.

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A root of px^{2} + qx + r = 0 is thrice the other root.Show that 3q^{2} = 16pr.

Solution of Example 2 of The Quadratic Formula :

The given equation is px^{2} + qx + r = 0 Comparing this equation with ax^{2} + bx + c = 0, we get a = p, b = q and c = r By data one root is thrice the other root. ⇒ If one root is α, then the other root is 3α α + 3α = -b⁄a = -q⁄p ⇒ 4α = -q⁄p ⇒ α = -q⁄4p......(i) α x 3α = c⁄a ⇒ 3α^{2} = r⁄p......(ii) Using (i) in (ii), we get 3(-q⁄4p)^{2} = r⁄p ⇒ 3q^{2}⁄16p^{2} = r⁄p Multiplying both sides with 16p^{2}, we get 3q^{2} = (16p^{2})(r⁄p) = 16pr (Proved.)

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Solve the following problems on The Quadratic Formula.

If α, β are the roots of x^{2} - (k + 1)x + (k^{2} + k + 1)⁄2 = 0, then show that α^{2} + β^{2} = k

For what values of k does the equation (k - 2)x^{2} + 2(2k - 3)x + (5k - 6) = 0have equal roots? For those values of k, find those equal roots.

For Answers See at the bottom of the Page.

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