# Theory of Equations – Problems I

### Theory of Equations – Problems I

1. Form polynomial equation of the lowest degree with roots as given below

a) 1, -1, 3

Solution: Equation having roots α, β, γ is (x – α) (x – β) (x – γ) = 0

Required equation is

(x – 1) (x + 1) (x – 3) = 0

(x² – 1) (x – 3) = 0

x³ – 3x² – x + 3 = 0

b) 1 ± 2i, 4, 2

Solution: Equation having roots α, β, γ is (x – α) (x – β) (x – γ) = 0

Required equation is

[x – (1 + 2i)] [x – (1 – 2i)] (x – 4) (x – 2) = 0

[x – (1 + 2i)] [x – (1 – 2i)]

= [(x – 1) – 2i] [(x – 1) + 2i]

= (x – 1)² – 4i²

= (x – 1)² + 4

= x² – 2x + 1 + 4

= x² – 2x + 5

(x – 4) (x – 2)

= x² – 4x – 2x + 8

= x² – 6x + 8

Required Equation (x² – 2x + 5) (x² – 6x + 8) = 0

⇒ x⁴ – 2x³ + 5x² – 6x³ + 12x² – 30x + 8x² – 16x + 40 = 0

⇒ x⁴ – 8x³ + 25x² – 46x + 40 = 0

2. If α, β, γ are the roots of 4x³ – 6x² + 7x + 3 = 0, then find the value of αβ + βγ + γα.

Solution: α, β, γ are the roots of 4x³ – 6x² + 7x + 3 = 0

α + β + γ = $$\frac{{{a}_{1}}}{{{a}_{0}}}=\frac{6}{4}$$,

αβ + βγ + γα = $$\frac{{{a}_{2}}}{{{a}_{0}}}=\frac{7}{4}$$,

αβγ = $$-\frac{{{a}_{3}}}{{{a}_{0}}}=\,-\frac{3}{4}$$,

∴ αβ + βγ + γα = $$\frac{7}{4}$$.

3. If 1, 1, α are the roots of x³ – 6x² + 9x – 4 = 0, then find α.

Solution: 1, 1, α are the roots of x³ – 6x² + 9x – 4 = 0

Sum = 1 + 1 + α = 6

α = 6 – 2 = 4.