THE QUADRATIC FORMULA - SOLVED AND EXERCISE PROBLEMS ON RELATION BETWEEN ROOTS AND COEFFICIENTS

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Please study
The Quadratic Formula
if you have not already done so.

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 x2 - px + q = 0, then find the value of

  1. α3 + β3
  2. 2⁄β) + (β2⁄α)
  3. (1⁄α3) + (1⁄β3)
  4. (α⁄β) + (β⁄α)


Solution of Example 1 of The Quadratic Formula :

The given equation is x2 - px + q = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -p and c = q
α, β are the roots of x2 - px + q = 0
We know, Sum of the roots = α + β = -ba = -(-p)⁄1 = p
Product of the roots = αβ = ca = q ⁄1 = q

(i) To find the value of α3 + β3

We know from Algebra Formula 6 in Algebra Formulas :

Cube of Sum of Two Terms:

(a + b)3 = a3 + 3ab(a + b) + b3

a3 + b3 = (a + b)3 - 3ab(a + b)
Applying this here, we get
α3 + β3 = (α + β)3 - 3αβ( α + β)
Using the values of α + β (= p) and αβ (= q) here, we get
α3 + β3 = ( p)3 - 3q(p) = p3 - 3pq Ans.

(ii) To find the value of (α2⁄β) + (β2⁄α)

Let A = (α2⁄β) + (β2⁄α)
Multiplying both sides with αβ, we get
αβA = αβ(α2⁄β) + αβ(β2⁄α)= α3 + β3
Using the values of α3 + β3 (= p3 - 3pq) and αβ (= q) here, we get
qA = p3 - 3pq ⇒ A = (p3 - 3pq)⁄q⇒ (α2⁄β) + (β2⁄α) = (p3 - 3pq)⁄q. Ans.

(iii) To find the value of (1⁄α3) + (1⁄β3)

Let A = (1⁄α3) + (1⁄β3)
Multiplying both sides with α3β3, we get
α3β3A = α3β3(1⁄α3) + α3β3 (1⁄β3)
(αβ)3A = β3 + α3 = α3 + β3
Using the values of α3 + β3 (= p3 - 3pq) and αβ (= q) here, we get
q3A = p3 - 3pq ⇒ A = (p3 - 3pq)⁄q3⇒ (1⁄α3) + (1⁄β3) = (p3 - 3pq)⁄q3. 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)= α2 + β2 = (α + β)2 - 2αβ
Using the values of α + β (= p) and αβ (= q) here, we get
qA = (p)2 - 2q ⇒ A = (p2 - 2q)⁄q
⇒ (α⁄β) + (β⁄α) = (p2 - 2q)⁄q. Ans.

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

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Example 2 : The Quadratic Formula

Solved Example 2 of The Quadratic Formula :

A root of px2 + qx + r = 0 is thrice the other root.Show that 3q2 = 16pr.

Solution of Example 2 of The Quadratic Formula :

The given equation is px2 + qx + r = 0
Comparing this equation with ax2 + 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 = -ba = -qp
⇒ 4α = -qp ⇒ α = -q⁄4p......(i)
α x 3α = ca ⇒ 3α2 = rp......(ii)
Using (i) in (ii), we get
3(-q⁄4p)2 = rp ⇒ 3q2⁄16p2 = rp
Multiplying both sides with 16p2, we get
3q2 = (16p2)(rp) = 16pr (Proved.)

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Exercise : The Quadratic Formula

Solve the following problems on The Quadratic Formula.

  1. If α, β are the roots of x2 - (k + 1)x + (k2 + k + 1)⁄2 = 0, then show that α2 + β2 = k
  2. For what values of k does the equation
    (k - 2)x2 + 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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Answers to Exercise : The Quadratic Formula

Answers to the problems on The Quadratic Formula :

  1. Proof
  2. k = 1, roots = -1,-1; k = 3, roots = -3,-3;