You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6

Question 1.
Discuss the maximum possible number of positive and negative roots of the polynomial equation 9x9 – 4x8 + 4x7 – 3x6 + 2x5 + x3 + 7x2 + 7x + 2 = 0.
Solution:
P(x) = 9x9 – 4x8 + 4x7 – 3x6 + 2x5 + x3 + 7x2 + 7x + 2
The number of sign changes in P(x) is 4.
∴P(x) has atmost 4 positive roots.
P(-x) = -9x9 – 4x8 – 4x7 – 3x6 – 2x5 – x3 + 7x2 – 7x + 2
The number of sign changes in P(-x) is 3.
P(x) has almost 3 negative roots. Since the difference between the number of sign changes in co-efficient P(-x) and the number of negative roots of the polynomial P(x) is even.
The number of negative roots = at most 2.

Question 2.
Discuss the maximum possible number of positive and negative zeros of the polynomials x2 – 5x + 6 and x2 – 5x + 16. Also, draw a rough sketch of the graphs.
Solution:
P(x) = x2 – 5x + 6
The number of sign changes in P(x) is 2.
P(x) has atmost 2 positive roots. P(-x) = x2 + 5x + 6.
The number of sign changes in P(-x) is 0.
∴ P (x) has no negative roots. P (x) = x2 – 5x + 16
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6 Q2

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6

Question 3.
Show that the equation x9 – 5x5 + 4x4 + 2x2 +1 = 0 has atleast 6 imaginary solutions.
Solution:
P(x) = x9 – 5x5 + 4x4 + 2x2 + 1
(i) The number of sign changes in P(x) is 2. The number of positive roots is atmost 2.
(ii) P(-x) = -x9 + 5x5 + 4x4 + 2x2 + 1. The number of sign changes in P(-x) is 1. The number of negative roots of P (x) is atmost 1. Since the difference of number of sign changes in P(-x) and number of negative zeros is even.
P(x) has one negative root.
(iii) 0 is not the zero of the polynomial P(x). So the number of real roots is almost 3.
∴ The number of imaginary roots at least 6.

Question 4.
Determine the number of positive and negative roots of the equation x9 – 5x8 – 14x7 = 0.
Solution:
x9 – 5x8 – 14x7 = 0
P (x) = x9 – 5x8 – 14x7. The number of sign changes is P(x) is 1.
The number of positive roots is 1. P (-x) = -x9 – 5x8 + 14x7
The number of sign changes is P(-x) is one. The number of negative zero of P(-x) is 1. It is clear that 0 is a root of the equation.
∴ The number of the imaginary roots is at least 6.

Question 5.
Find the exact number of real zeros and imaginary of the polynomial x9 + 9x7 + 7x5 + 5x3 + 3x.
Solution:
P(x) = x9 + 9x7 + 7x5 + 5x3 + 3x.
There is no change in the sign of P(x) and P(-x), P(x) has no positive and no negative real roots, but 0 is the root of the polynomial equation P(x).

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6 Additional Problems

Question 1.
Find the maximum possible number of real roots of the equation, x5 – 6.x2 – 4x + 5 = 0.
Solution:
Let f(x) = x5 – 6x2 – 4x + 5
Check the terms when it changes sign
Number of changes = 2
∴ Maximum number of positive real roots = 2
f(-x) = (- x)5 – 6 (- x)2 – 4 (-x) + 5
= -x5 – 6x2 + 4x + 5
Check the terms when it changes signs.
Number of changes = 1
∴ Maximum number of negative real roots = 1
∴ Total maximum number of real roots = 2 + 1 = 3

Question 2.
Find the number of real roots of the equation, x2 + 5 |x| + 6 = 0.
Solution:
|x2| – 5|x| + 6 = 0
(|x| -2) (|x| -3) = 0
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6 1
It has four real roots. The real roots are 2, -2, 3, -3

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6

Question 3.
Find the real roots of the equation, x2 + 5 |x| + 6 = 0.
Solution:
x2 + 5 |x| + 6 = 0
Case (i) If x ≥ 0
x2 + 5x + 6 = 0
(x + 2) (x + 3) = 0
x = -2 and x = -3
Case (ii) If x < 0
x2 – 5x + 6 = 0
(x – 2) (x – 3) = 0
x = 2 and x = 3

Question 4.
Solve x4 – 4x2 + 8x + 35 = 0. Given (2 + \(i \sqrt{3}\)) is a root.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6 5
⇒ x2 – 4x + 7 is a factor of P (x). Dividing the polynomial P(x) = 0 by x2 – 4x + 7.
We get x2 + 4x + 5 = 0 is a other factor. The roots of x2 + 4x + 5 = 0 are
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6 6

Question 5.
Solve x4 – 5x3 + 4x2 + 8x – 8 = 0. Given (1 – \(\sqrt{5}\)) is a root of the polynomial equation.
Solution:
Since (1 – \(\sqrt{5}\)) is a root of the polynomial P(x) = 0
(1 + \(\sqrt{5}\)) is also a root of P (x) = 0
⇒ x2 – [(1 + \(\sqrt{5}\)) + (1 – \(\sqrt{5}\))] x + (1 + \(\sqrt{5}\) ) (1 – \(\sqrt{5}\) ) = 0 is a factor of P(x) = 0 ⇒ x2 – 2x – 4 = 0 is a factor of P(x) = 0.
Dividing the polynomial by x2 – 2x – 4 = 0
We get the other factor x2 – 3x + 2 = 0
The roots of x2 – 3x + 2 = 0
(x – 2) (x – 1) = 0
x = 1, 2
The roots are 1, 2, 1 ± \(\sqrt{5}\)

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6

Question 6.
Find a polynomial equation of the lowest degree with rational co-efficient having \(\sqrt{3}\), (1 – 2i) as two of its roots.
Solution:
When \(\sqrt{3}\) is a root, – \(\sqrt{3}\) will also be a root.
Now the quadratic equation with \(\sqrt{3}\) , – \(\sqrt{3}\) are roots is x2 – (\(\sqrt{3}\) – \(\sqrt{3}\))x + (\(\sqrt{3}\))(-\(\sqrt{3}\)) = 0
(i.e) x2 – 3 = 0
When 1 – 2i is a root, 1 + 2i will be another root.
Now the quadratic equation with roots 1 – 2i and 1 + 2i is
x2 – (1 – 2i + 1 + 2i)x + (1 – 2i)(1 + 2i) = 0
(i.e) x2 – 2x + 5 = 0
∴ The equation with roots ± \(\sqrt{3}\) and 1 ± 2i is (x2 – 3) (x2 – 2x + 5) = 0
(i.e) x4 – 2x3 + 5x2 – 3x2 + 6x – 15 = 0 ,
(i.e) x4 – 2x3 + 2x2 + 6x – 15 = 0

Leave a Reply

Your email address will not be published. Required fields are marked *