PROBLEMS INVOLVING SUM AND PRODUCT OF ROOTS WITH UNKNOWNS

Problem 1 :

Find the sum and product of the roots of :

3x² – 2x + 7 = 0

Solution :

3x² - 2x + 7 = 0

a = 3, b = -2, c = 7

Problem 2 :

Find the sum and product of the roots of :

x² + 11x - 13 = 0

Solution :

x² + 11x - 13 = 0

a = 1, b = 11, c = -13

Problem 3 :

Find the sum and product of the roots of :

 5x² – 6x - 14 = 0

Solution :

5x² - 6x - 14 = 0

a = 5, b = -6, c = -14

Problem 4 :

The equation kx² – (1 + k)x + (3k + 2) = 0 is such that the sum of its roots is twice their product. Find k and the two roots.

Solution :

kx² – (1 + k)x + (3k + 2) = 0

Let α and β be two roots.

α + β = 2 (α β)

a = k, b = -(1 + k), c = 3k + 2

The sum of its roots is twice their product.

α + β = 2(α β)

So, (1 + k)/k = 2(3k + 2)/k

Multiplying each side by k.

k ⋅ (1 + k)/k = 2(3k + 2)/k ⋅ k

1 + k = 2(3k + 2)

1 + k = 6k + 4

1 – 4 = 6k – k

-3 = 5k

-3/5 = k

k = -3/5 substitute in given equation.

kx² – (1 + k)x + (3k + 2) = 0

(-3/5)x² – (1 + (-3/5))x + (3(-3/5) + 2) = 0

-3/5x² – (1 – 3/5)/x + (-9/5 + 2) = 0

-3/5x² – ((5 – 3)/5)/x + (-9 + 10)/5 = 0

-3/5x² – 2/5x + 1/5 = 0

Dividing each side by 5.

-3x² - 2x + 1 = 0

-(3x² + 2x – 1) = 0

3x² + 2x – 1 = 0

(3x – 1) (x + 1) = 0

 3x – 1 = 0 and x + 1 = 0

3x = 1 and x = -1

x = 1/3

So, two roots are 1/3 and -1.

Problem 5 :

The quadratic equation ax² – 6x + a - 2 = 0, a ≠ 0, has one root which is double the other.

a) Let the roots be α and 2α. Hence find two equations involving α.

b)  Find α and the two roots of the quadratic equation.

Solution :

a) To find two equations involving α :

ax² – 6x + a - 2 = 0

a = a, b = -6, c = a – 2

α = α, β = 2α

b) By applying the value of α in (2), we get

(2/a)² = (a – 2)/2a

4/a² = (a – 2)/2a

8 = a(a – 2)

8 = a² – 2a

8 = a² – 2a

a² – 2a – 8 = 0

(a + 2) (a – 4) = 0

So, a = -2 and a = 4

If a = -2, the roots are -1 and -2.

If a = 4, the roots are 1/2 and 1.