PROBLEMS WITHE RELATIONSHIP BETWEEN ROOTS OF A QUADRATIC EQUATION

Problem 1 :

The quadratic equation kx² + (k–8)x + (1 – k) = 0, k ≠ 0, has one root which is two more than the other. Find k and the two roots.

Solution :

Let α and β be the roots of the equation.

α = β + 2

α + β  = -(k - 8)/k ==> (8 - k)/k

β + 2 + β = (8 - k)/k

2(β + 1) = (8 - k)/k

α β  = (1 - k)/k

(k - 16)(k - 4) = 0

k = 16 and k = 4

Problem 2 :

The roots of the equation x² – 6x + 7 = 0 are α and β. Find the simplest quadratic equation with roots

α + 1/β and β + 1/α.

Solution :

x² – 6x + 7 = 0

a = 1, b = -6, c = 7

α = α + 1/β, β = β + 1/α

Sum of the roots = α + β = -b/a ==> 6

Products of the roots = αβ = c/a ==> 7

Sum of roots :

= α + 1/β + β + 1/α

= (α + β) + (1/α + 1/β)

= (α + β) + ((α + β)/αβ)

= 6 + (6/7)

= 48/7

Product of roots :

= (α + 1/β) (β + 1/α)

= αβ + 1 + 1 + 1/αβ

= αβ + 2 + 1/αβ

= 7 + 2 + (1/7)

= 9 + (1/7)

= 64/7

x² – (48/7)x + (64/7) = 0

7x² – 48x + 64 = 0

Problem 3 :

The roots of 2x² – 3x - 5 = 0 are p and q. Find all quadratic equations with roots p² + q and q² + p.

Solution :

From the given quadratic equation :

2x² – 3x - 5 = 0

p and q are roots.

p + q = 3/2 and pq = -5/2

Sum and product of the roots of the required quadratic equation.

α = p² + q and β = q² + p

Sum of the roots :

α + β = p² + q + q² + p

= p² + q² + p + q

Algebraic identity for a² + b² = (a+b)²- 2ab

Algebraic identity for a³ + b³ = (a+b)³- 3ab(a + b)

= (3/2)² - 2(-5/2) + (3/2)

= (9/4) + 5 + 3/2

= (9 + 20 + 6)/4

= 35/4

Product of the roots :

α β = (p² + q) (q² + p)

= p² q² + p³ + q³ + pq

= (pq)² + p³ + q³ + pq

= (-5/2)² + (3/2)³ - 3(-5/2)(3/2) + (-5/2)

= (25/4) + (27/8) + (45/4) + (-5/2)

= (70/4) + (27/8) + (-5/2)

= 147/8

x² - (35x/4) + (147/8) = 0

8x² - 70x + 147 = 0

So, the required equation is a(8x² - 70x + 147) = 0

Problem 4 :

kx² + (k + 2) x - 3 = 0

has roots which are real and positive. Find the possible values that k may have.

Solution :

Since the roots are real and positive, we may use the condition

b2 - 4ac > 0

a = k, b = k+2 and c = -3

(k+2)² - 4k(-3) = 0

k² + 4k + 4 + 12k = 0

k² + 16k + 4 = 0

To solve for k, we use quadratic formula.

-8±√60 ≤ k < 0