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

if you have not already done so.

There, We presented the Derivation Of
Formulae Relating roots and coefficients
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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