Solve the equation 3 x 3 - 4 x 2 + x + 88 = 0 which has 2 - - 7 as a root.

Given that, 2 - 7 i is one root. ⇒ 2 + 7 i is another root. We know that, the equation having roots α , β is x 2 - α + β x + α β = 0 The equation having roots 2 ± 7 i is x 2 - 2 - 7 i + 2 + 7 i x + 2 - 7 i 2 + 7 i = 0 ⇒ x 2 - 4 x + 11 = 0 . ∴ x 2 - 4 x + 11 = 0 is a factor of 3 x 3 - 4 x 2 + x + 88 = 0 . Applying synthetic division. 4 - 11 3 0 0 - 4 12 0 1 32 - 33 88 0 - 88 3 8 0 0 ⇒ 3 x + 8 = 0 ⇒ x - 8 3 Therefore, the roots of the given equation are 2 ± 7 i , - 8 3 .

Solve x 4 - 4 x 2 + 8 x + 35 = 0 , given that 2 + i 3 is a root.

Given that, 2 + i 3 is one root. ⇒ 2 - i 3 is another root. We know that, the equation having roots α , β is x 2 - α + β x + α β = 0 The equation having roots 2 ± i 3 is x 2 - 2 + i 3 + 2 - i 3 x + 2 + i 3 2 - i 3 = 0 ⇒ x 2 - 4 x + 7 = 0 . ∴ x 2 - 4 x + 7 = 0 is a factor of x 4 - 4 x 2 + 8 x + 35 = 0 . 4 - 7 1 0 0 0 4 0 - 4 16 - 7 8 20 - 28 35 0 - 35 1 4 5 0 0 ⇒ x 2 + 4 x + 5 = 0 Now, applying quadratic formula. ⇒ x = - 4 ± 16 - 20 2 x = - 4 ± 2 i 2 = - 2 ± i Therefore, the roots of the given equation are 2 ± i 3 , - 2 ± i .

Solve the equation x 4 - 6 x 3 + 11 x 2 - 10 x + 2 = 0 , given that 2 + 3 is a root of the equation.

Given that, 2 - 3 i is one root. ⇒ 2 + 3 i is another root. We know that, the equation having roots α , β is x 2 - α + β x + α β = 0 The equation having roots 2 ± i 3 is x 2 - 2 + 3 i + 2 - 3 i x + 2 + 3 i 2 + 3 i = 0 ⇒ x 2 - 4 x + 1 = 0 . ∴ x 2 - 4 x + 1 = 0 is a factor of x 4 - 6 x 3 + 11 x 2 - 10 x + 2 = 0 . 4 - 1 1 0 0 - 6 4 0 11 - 8 - 1 - 10 8 2 2 0 - 2 1 - 2 2 0 0 ⇒ x 2 - 2 x + 2 = 0 Now, applying quadratic formula. ⇒ x = 2 ± 4 - 8 2 = - 2 ± 2 i 2 ⇒ x = - 1 ± i Therefore, the roots of the given equation are 2 ± 3 , 1 ± i .

Given that - 2 + - 7 is a root of the equation x 4 + 2 x 2 - 16 x + 77 = 0 , solve it completely.

Given that, - 2 - 7 i is one root. ⇒ - 2 + 7 i is another root. The equation having roots - 2 ± 7 i is x 2 - - 2 - 7 i + - 2 + 7 i x + - 2 - 7 i - 2 + 7 i = 0 ⇒ x 2 + 4 x + 11 = 0 . ∴ x 2 + 4 x + 11 = 0 is a factor of x 4 + 2 x 2 - 16 x + 77 = 0 . Now, applying synthetic division - 4 - 11 1 0 0 0 - 4 0 2 16 - 11 - 16 - 28 44 77 0 - 77 1 - 4 7 0 0 ⇒ x 2 - 4 x + 7 = 0 Now, applying quadratic formula ⇒ x = 4 ± 16 - 28 2 = 4 ± 2 3 i 2 = 2 ± 3 i Therefore, the roots of the given equation are - 2 ± i 7 , 2 ± 3 i .

Solve the equation x 4 + 2 x 3 - 16 x 2 - 22 x + 7 = 0 , given that 2 - 3 is one of its roots.

Let 2 - 3 is one root. If a - i b is a root then a + i b is also a root. ⇒ 2 + 3 is another root. The equation having roots α , β is given by x 2 - α + β x + α β = 0 The equation having roots 2 ± 3 is x 2 - 2 - 3 + 2 + 3 x + 2 - 3 2 + 3 = 0 ⇒ x 2 - 4 x + 1 = 0 . ∴ x 2 - 4 x + 1 = 0 is a factor of x 4 + 2 x 3 - 16 x 2 - 22 x + 7 = 0 . Now, applying synthetic division 4 - 1 1 0 0 2 4 0 - 16 24 - 1 - 22 28 - 6 7 0 - 7 1 6 7 0 0 ⇒ x 2 + 6 x + 7 = 0 On applying quadratic formula ⇒ x = - 6 ± 36 - 28 2 x = - 6 ± 8 2 = - 3 ± 2 Therefore, the roots of the given equation are 2 ± 3 , - 3 ± 2 .

Solve the equation 3 x 5 - 4 x 4 - 42 x 3 + 56 x 2 + 27 x - 36 = 0 given that 2 + 5 is one of its roots.

Let 2 + 5 is one root. ⇒ 2 - 5 is another root. The equation having roots 2 ± 5 is x 2 - ( 2 + 5 + 2 - 5 ) x + ( 2 + 5 ) ( 2 - 5 ) = 0 ⇒ x 2 - 2 2 x - 3 = 0 The equation having roots ± 2 ± 5 is x 2 + 2 2 x - 3 x 2 - 2 2 x - 3 = 0 ⇒ x 2 - 3 + 2 2 x 2 - 3 - 2 2 = 0 ⇒ x 2 - 3 2 - ( 2 2 x ) 2 = 0 ⇒ x 4 - 6 x 2 + 9 - 8 x 2 = 0 ⇒ x 4 - 14 x 2 + 9 = 0 Given 3 x 5 - 4 x 4 - 42 x 3 + 56 x 2 + 27 x - 36 = 0 ⇒ 3 x x 4 - 14 x 2 + 9 - 4 x 4 - 14 x 2 + 9 = 0 ⇒ x 4 - 14 x 2 + 9 ( 3 x - 4 ) = 0 ⇒ x = ± 2 ± 5 or x = 4 3 Therefore, the roots of the given equation are ± 2 ± 5 , 4 3 .

Show that the equation a 2 x - a ' + b 2 x - b + c 2 x - c 2 + … + k 2 x - k ' = x - m , where a , b , c , … , k , m , a ' , b j , … , k ' are all real numbers, can not have a non-real root.

a 2 x - a ' + b 2 x - b ' + c 2 x - c ' + … + k 2 x - k ' = x - m Let α + i β be a root of the equation. If β ≠ 0 , α - i β is also a root of the equation. Substitute α + i β in the equation. a 2 α + i β - a ' + b 2 α + i β - b ' + c 2 α + i β - c ' + … … + k 2 α - i β - k ' = α + i β - m ⇒ a 2 α - a ' - i β α - a ' 2 + β 2 + b 2 α - b ' - i β α - b ' 2 + β 2 + . . . . + k 2 α - k ' - i β α - k ' 2 - β 2 = α + i β - m ....(i) Substitute α - i β in the equation ⇒ a 2 α - a ' + i β α - a ' 2 + β 2 + b 2 α - b ' + i β α - b ' 2 + β 2 + … + k 2 α - k ' + i β α - k ' 2 + β 2 = α - i β - m ......(ii) (ii) - (i) gives 2 i β . ⇒ a 2 α - a ' 2 + β 2 + b 2 α - b ' 2 + β 2 + … . + k 2 α - k ' 2 - β 2 = - 2 i β ⇒ 2 i β ⇒ a 2 α - a ' 2 + β 2 + b 2 α - b ' 2 + β 2 + … + k 2 α - k ' 2 - β 2 + 1 = 0 ⇒ β = 0 This is a contradiction. ∴ The given equation can not have non real roots.

E M B I B E

Intermediate Second Year Mathematics Paper 2A>Algebra>Chapter 4 - Theory of Equations>Exercise 4(c)

Andhra Pradesh Board Solutions for Exercise 3: Exercise 4(c)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 3: Exercise 4(c)

Attempt the free practice questions from Exercise 3: Exercise 4(c) with hints and solutions to strengthen your understanding. Intermediate Second Year Mathematics Paper 2A solutions are prepared by Experienced Embibe Experts.

Questions from Andhra Pradesh Board Solutions for Exercise 3: Exercise 4(c) with Hints & Solutions

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 4 - Theory of Equations>Exercise 4(c)>Q II

Solve the equation $3 x^{3} - 4 x^{2} + x + 88 = 0$ which has $2 - \sqrt{- 7}$ as a root.

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 4 - Theory of Equations>Exercise 4(c)>Q II

Solve $x^{4} - 4 x^{2} + 8 x + 35 = 0,$ given that $2 + i \sqrt{3}$ is a root.

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 4 - Theory of Equations>Exercise 4(c)>Q II

Solve the equation $x^{4} - 6 x^{3} + 11 x^{2} - 10 x + 2 = 0$ , given that $2 + \sqrt{3}$ is a root of the equation.

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 4 - Theory of Equations>Exercise 4(c)>Q II

Given that $- 2 + \sqrt{- 7}$ is a root of the equation $x^{4} + 2 x^{2} - 16 x + 77 = 0$ , solve it completely.

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 4 - Theory of Equations>Exercise 4(c)>Q II

Solve the equation $x^{4} + 2 x^{3} - 16 x^{2} - 22 x + 7 = 0$ , given that $2 - \sqrt{3}$ is one of its roots.

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 4 - Theory of Equations>Exercise 4(c)>Q II

Solve the equation $3 x^{5} - 4 x^{4} - 42 x^{3} + 56 x^{2} + 27 x - 36 = 0$ given that $\sqrt{2} + \sqrt{5}$ is one of its roots.

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 4 - Theory of Equations>Exercise 4(c)>Q II

Solve the equation $x^{4} - 9 x^{3} + 27 x^{2} - 29 x + 6 = 0$ , given that one root is $2 - \sqrt{3}$ .

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 4 - Theory of Equations>Exercise 4(c)>Q II

Show that the equation $\frac{a^{2}}{x - a^{'}} + \frac{b^{2}}{x - b} + \frac{c^{2}}{x - c^{2}} + \dots + \frac{k^{2}}{x - k^{'}} = x - m,$ where $a, b, c, \dots, k, m, a^{'}, b_{j}, \dots, k^{'}$ are all real numbers, can not have a non-real root.

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