## Rajasthan Board RBSE Class 11 Maths Chapter 5 Complex Numbers Ex 5.4

Question 1.

Find the solution of following equations by vedic method

(i) x^{2} + 4x + 13 = 0

(ii) 2x^{2} + 5x + 4 = 0

(iii) ix^{2} + 4x – 15/2 = 0

Solution:

(i) Given equation

x^{2} + 4x + 13 = 0

Comparing this equation with ax^{2} + bx + c = 0, we get

a = 1, b = 4, c = 13

First derivative = Discriminant

Question 2.

Find quadratic equation which has the following roots

(i) 5 and -2

(ii) 1 + 2i

Solution:

(i) root α = 5 and β = – 2

Then, sum of roots = α + β = 5 – 2

⇒ α + β = 3

and product of roots = αβ = 5 × (-2)

⇒ αβ = -10

Hence, required equation whose roots are 5 and -2.

x^{2} – (sum of roots) x + product of roots = 0

⇒ x^{2} – 3x + (-10) = 0

⇒ x^{2} – 3x – 10 = 0

Hence, required equation is x^{2} – 3x – 10 = 0 whose roots are 5 and -2.

(ii) Roots α = 1 + 2i and β = 1 – 2i

Then, sum of roots α + β = 1 + 2i + 1 – 2i = 2

and product of roots αβ = (1 + 2i)(1 – 2i) = 1 – 4i^{2} = 1 + 4 = 5

Hence, required equation whose roots are 1 + 2i and 1 – 2i,

x^{2} – (sum of roots) x + Product of roots = 0

⇒ x^{2} – 2x + 5 = 0.

Remark: If one root is 1 + 2i, then second not will be 1 – 2i.

Question 3.

If one root of equation x^{2} – px + q = 0 is twice the other then prove that 2p^{2} = 9q.

Solution:

Given equation

x^{2} – px + q = 0

Let its roots be α and β, then

According to question α = 2β then

Question 4.

Find that condition for which equation ax^{2} + bx + c = 0 has roots in the ratio m : n.

Solution:

Given equation,

ax^{2} + bx + c = 0

Let its roots be α and β, then

According to question,

α : β = m : n

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