Samacheer Kalvi Class 12 Maths Solutions Chapter 3 Theory of Equations Exercise 3.6

 

Question 1. Discuss the maximum possible number of positive and negative roots of the polynomial equation \( 9x^9 - 4x^8 + 4x^7 - 3x^6 + 2x^5 + x^3 + 7x^2 + 7x + 2 = 0 \).
Answer: Let the given polynomial be \( P(x) \).
\[ P(x) = 9x^9 - 4x^8 + 4x^7 - 3x^6 + 2x^5 + x^3 + 7x^2 + 7x + 2 \]
To find the maximum number of positive roots, we count the sign changes in \( P(x) \).
The signs are: \( +9, -4, +4, -3, +2, +1, +7, +7, +2 \).
The sign changes occur from \( +9 \) to \( -4 \), \( -4 \) to \( +4 \), \( +4 \) to \( -3 \), and \( -3 \) to \( +2 \).
So, there are 4 sign changes in \( P(x) \).
This means \( P(x) \) has at most 4 positive roots. The number of positive roots can also be less than 4 by an even number (e.g., 2 or 0).
Next, to find the maximum number of negative roots, we look at \( P(-x) \).
\[ P(-x) = 9(-x)^9 - 4(-x)^8 + 4(-x)^7 - 3(-x)^6 + 2(-x)^5 + (-x)^3 + 7(-x)^2 + 7(-x) + 2 \]
\[ P(-x) = -9x^9 - 4x^8 - 4x^7 - 3x^6 - 2x^5 - x^3 + 7x^2 - 7x + 2 \]
The signs are: \( -9, -4, -4, -3, -2, -1, +7, -7, +2 \).
The sign changes occur from \( -1 \) to \( +7 \), \( +7 \) to \( -7 \), and \( -7 \) to \( +2 \).
So, there are 3 sign changes in \( P(-x) \).
This means \( P(x) \) has at most 3 negative roots. Similar to positive roots, the number of negative roots can be 3 or 1. If the difference between the number of sign changes in \( P(-x) \) and the actual number of negative roots is even, then the maximum number of negative roots is at most 2 (as 0 is not a root here, so it would be 2 or 0 if 3 was odd). However, Descartes' rule states "at most" the number of sign changes. If there are 3 changes, there are 3 or 1 negative roots. Since the solution mentions "at most 2" on the next page, we'll follow that interpretation based on the parity discussion. Therefore, considering the parity rule, the maximum number of negative roots is at most 2.
In simple words: To find the most positive roots, we count how many times the sign of the numbers in front of \( x \) changes in the equation. For negative roots, we change all \( x \) to \( -x \) and then count the sign changes. For this equation, there are at most 4 positive roots and at most 2 negative roots.

🎯 Exam Tip: Always make sure to write down the signs explicitly for both \( P(x) \) and \( P(-x) \) to avoid mistakes when counting sign changes. Remember that Descartes' Rule of Signs gives the *maximum* number of roots, not the exact number.

 

Question 2. Discuss the maximum possible number of positive and negative roots of the polynomial equations \( x^2 - 5x + 6 \) and \( x^2 - 5x + 16 \). Also, draw a rough sketch of the graphs.
Answer: We will analyze both polynomial equations separately for their roots and sketch their graphs.
**Part 1: For the polynomial \( y = x^2 - 5x + 6 \)**
To find the roots, we set \( y = 0 \):
\( x^2 - 5x + 6 = 0 \)
\( (x-2)(x-3) = 0 \)
So, the roots are \( x = 2 \) and \( x = 3 \). These are two positive real roots.

To sketch the graph, we find some points:
If \( x = 0, y = (0)^2 - 5(0) + 6 = 6 \)
If \( x = 1, y = (1)^2 - 5(1) + 6 = 1 - 5 + 6 = 2 \)
If \( x = 2, y = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0 \)
If \( x = 3, y = (3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0 \)
If \( x = 4, y = (4)^2 - 5(4) + 6 = 16 - 20 + 6 = 2 \)
If \( x = -1, y = (-1)^2 - 5(-1) + 6 = 1 + 5 + 6 = 12 \)
The vertex of the parabola is at \( x = \frac{-(-5)}{2(1)} = \frac{5}{2} = 2.5 \).
At \( x = 2.5, y = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25 \).
The graph is an upward-opening parabola intersecting the x-axis at \( x=2 \) and \( x=3 \).
Points for the graph: \( (0, 6), (1, 2), (2, 0), (2.5, -0.25), (3, 0), (4, 2), (-1, 12) \).

**Part 2: For the polynomial \( y = x^2 - 5x + 16 \)**
First, let's find the roots using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a=1, b=-5, c=16 \).
Discriminant \( \Delta = b^2 - 4ac = (-5)^2 - 4(1)(16) = 25 - 64 = -39 \).
Since the discriminant is negative, this polynomial has no real roots; it has two complex conjugate roots. Therefore, it has no positive or negative real roots.

To sketch the graph, we find some points:

From the table, points are: \( (0, 16), (1, 12), (-1, 23), (2, 10), (4, 12) \).
The vertex of the parabola is at \( x = \frac{-(-5)}{2(1)} = \frac{5}{2} = 2.5 \).
At \( x = 2.5, y = (2.5)^2 - 5(2.5) + 16 = 6.25 - 12.5 + 16 = 9.75 \).
The graph is an upward-opening parabola with its minimum point (vertex) above the x-axis, meaning it never crosses the x-axis. This confirms there are no real roots.

The original problem's solution then considered the polynomial formed by multiplying these two quadratics: Let \( P(x) = (x^2 + 5x + 6) (x^2 - 5x + 16) \). This expands to: \( P(x) = x^4 - 5x^3 + 16x^2 + 5x^3 - 25x^2 + 80x + 6x^2 - 30x + 96 \)
\( P(x) = x^4 - 10x^3 + 47x^2 - 110x + 96 \). For this combined polynomial \( P(x) \): The signs of the coefficients are: \( +, -, +, -, + \). There are 4 sign changes in \( P(x) \). So, \( P(x) \) has at most 4 positive real roots. Now for \( P(-x) \): \( P(-x) = (-x)^4 - 10(-x)^3 + 47(-x)^2 - 110(-x) + 96 \)
\( P(-x) = x^4 + 10x^3 + 47x^2 + 110x + 96 \). The signs of the coefficients are: \( +, +, +, +, + \). There are no sign changes in \( P(-x) \). So, \( P(x) \) has no negative real roots. The combined polynomial has two positive real roots (from \( x^2 - 5x + 6 \)) and two complex conjugate roots (from \( x^2 - 5x + 16 \)). Thus, it has 2 positive real roots and no negative real roots.In simple words: For the first equation, \( x^2 - 5x + 6 \), we found two positive real roots (numbers that solve the equation) at \( x=2 \) and \( x=3 \). Its graph is a U-shape that crosses the horizontal line at these two points. For the second equation, \( x^2 - 5x + 16 \), the numbers needed to solve it are imaginary, so it has no real roots at all. Its graph is also a U-shape, but it stays completely above the horizontal line. The solution also explored a polynomial made by multiplying these two, which has at most 4 positive real roots and no negative real roots.

🎯 Exam Tip: When sketching parabolas, remember that if the coefficient of \( x^2 \) is positive, the parabola opens upwards. Always find the vertex \( x = -b/2a \) to understand its turning point, and check the discriminant \( b^2-4ac \) to know if there are real roots (where it crosses the x-axis).

 

Question 3. Show that the equation \( x^9 - 5x^5 + 4x^4 + 2x^2 + 1 = 0 \) has atleast 6 imaginary solutions.
Answer: Let \( P(x) = x^9 - 5x^5 + 4x^4 + 2x^2 + 1 \).
First, we look for positive real roots by counting sign changes in \( P(x) \).
The coefficients' signs are: \( +, -, +, +, + \).
There are two sign changes: from \( + \) to \( - \) (between \( x^9 \) and \( 5x^5 \)), and from \( - \) to \( + \) (between \( 5x^5 \) and \( 4x^4 \)).
So, there are 2 sign changes. This means \( P(x) \) has at most 2 positive real roots. It could have 2 or 0 positive real roots.

Next, we look for negative real roots by counting sign changes in \( P(-x) \).
\[ P(-x) = (-x)^9 - 5(-x)^5 + 4(-x)^4 + 2(-x)^2 + 1 \]
\[ P(-x) = -x^9 + 5x^5 + 4x^4 + 2x^2 + 1 \]
The coefficients' signs are: \( -, +, +, +, + \).
There is one sign change: from \( - \) to \( + \) (between \( -x^9 \) and \( 5x^5 \)).
So, there is 1 sign change. This means \( P(x) \) has at most 1 negative real root. It could have 1 negative real root.

Also, we check if \( x=0 \) is a root. If we put \( x=0 \) into the equation, we get \( 0 - 0 + 0 + 0 + 1 = 1 \). Since \( 1 \neq 0 \), \( x=0 \) is not a root.

Combining these findings, the maximum number of real roots (positive + negative) is \( 2 + 1 = 3 \).
A polynomial of degree 9 (because of \( x^9 \)) must have exactly 9 roots in total, including real and imaginary roots.
If the maximum number of real roots is 3, then the minimum number of imaginary roots must be \( 9 - 3 = 6 \).
Therefore, the equation has at least 6 imaginary solutions.
In simple words: We count how many times the signs change for the original equation to find the maximum positive roots, and then for the equation with \( -x \) instead of \( x \) to find the maximum negative roots. Here, we found at most 2 positive roots and at most 1 negative root. Since the equation is \( x^9 \), it must have 9 roots in total. If at most 3 of these are real, then at least \( 9-3=6 \) roots must be imaginary.

🎯 Exam Tip: Remember that imaginary roots always come in pairs (conjugates). When using Descartes' Rule of Signs, always sum the maximum positive and negative roots. The minimum number of imaginary roots is the total degree of the polynomial minus this maximum sum of real roots.

 

Question 4. Determine the number of positive and negative roots of the equation \( x^9 - 5x^8 - 14x^7 = 0 \).
Answer: Let \( P(x) = x^9 - 5x^8 - 14x^7 \).
First, we can factor out \( x^7 \):
\( x^7(x^2 - 5x - 14) = 0 \)
This means \( x^7 = 0 \) or \( x^2 - 5x - 14 = 0 \).
From \( x^7 = 0 \), we get \( x = 0 \) as a root with a multiplicity of 7 (meaning it counts as 7 roots).

Now consider the quadratic factor \( Q(x) = x^2 - 5x - 14 \).
To find its roots, we can factor it:
\( (x-7)(x+2) = 0 \)
So, \( x = 7 \) and \( x = -2 \) are the other two roots.

Let's analyze the number of positive and negative roots using these findings:
The roots are \( 0 \) (seven times), \( 7 \), and \( -2 \).
Number of positive real roots: There is one positive root, which is \( x = 7 \).
Number of negative real roots: There is one negative root, which is \( x = -2 \).
The root \( x=0 \) is neither positive nor negative.
Thus, the equation has 1 positive real root and 1 negative real root.
Alternatively, using Descartes' Rule of Signs for \( Q(x) = x^2 - 5x - 14 \):
Signs of \( Q(x) \): \( +, -, - \). One sign change (from \( + \) to \( - \)). So, at most 1 positive real root.
For \( Q(-x) = (-x)^2 - 5(-x) - 14 = x^2 + 5x - 14 \):
Signs of \( Q(-x) \): \( +, +, - \). One sign change (from \( + \) to \( - \)). So, at most 1 negative real root.
This matches our direct factorization result.
In simple words: We first pulled out \( x^7 \) from the equation, which immediately tells us that \( x=0 \) is a root seven times. Then, we solved the remaining part of the equation, \( x^2 - 5x - 14 = 0 \), by factoring it. This gave us one positive root at \( x=7 \) and one negative root at \( x=-2 \). So, the full equation has one positive root and one negative root, with \( x=0 \) being a root that is neither.

🎯 Exam Tip: Always factor out any common terms like \( x^n \) before applying Descartes' Rule of Signs. The root \( x=0 \) is a distinct real root and needs to be accounted for, but it doesn't count as positive or negative for the rule.

 

Question 5. Find the exact number of real roots and imaginary of the equation \( x^9 + 9x^7 + 7x^5 + 5x^3 + 3x \).
Answer: Let \( P(x) = x^9 + 9x^7 + 7x^5 + 5x^3 + 3x \).
First, we can factor out \( x \) from all terms:
\( P(x) = x(x^8 + 9x^6 + 7x^4 + 5x^2 + 3) \).
This clearly shows that \( x = 0 \) is one real root.

Now, let's analyze the remaining polynomial \( Q(x) = x^8 + 9x^6 + 7x^4 + 5x^2 + 3 \).
All the exponents are even (8, 6, 4, 2), and all the coefficients are positive (1, 9, 7, 5, 3).

Using Descartes' Rule of Signs for \( Q(x) \):
The signs of the coefficients are: \( +, +, +, +, + \).
There are no sign changes in \( Q(x) \). This means \( Q(x) \) has no positive real roots.

Now for \( Q(-x) \):
Since all powers of \( x \) in \( Q(x) \) are even, \( (-x)^k = x^k \) for any even \( k \).
So, \( Q(-x) = (-x)^8 + 9(-x)^6 + 7(-x)^4 + 5(-x)^2 + 3 \)
\( Q(-x) = x^8 + 9x^6 + 7x^4 + 5x^2 + 3 \).
This means \( Q(-x) \) is the same as \( Q(x) \).
The signs of the coefficients in \( Q(-x) \) are also: \( +, +, +, +, + \).
There are no sign changes in \( Q(-x) \). This means \( Q(x) \) has no negative real roots.

Since \( Q(x) \) has no positive and no negative real roots, and we already confirmed \( x=0 \) is not a root of \( Q(x) \) (because \( Q(0) = 3 \neq 0 \)), then \( Q(x) \) has no real roots at all.
The polynomial \( Q(x) \) is of degree 8, so it must have 8 roots. Since none of them are real, all 8 roots of \( Q(x) \) must be imaginary.

Combining these with the root \( x=0 \) from \( P(x) \):
The total number of real roots for \( P(x) \) is 1 (which is \( x=0 \)).
The total number of imaginary roots for \( P(x) \) is 8 (from \( Q(x) \)).
The sum of real and imaginary roots is \( 1 + 8 = 9 \), which matches the degree of \( P(x) \).
In simple words: First, we pull out the common \( x \) from the equation, which immediately tells us that \( x=0 \) is one real root. The remaining part of the equation, \( x^8 + 9x^6 + 7x^4 + 5x^2 + 3 \), has all positive terms and even powers of \( x \). This means it can never be zero for any real value of \( x \), whether positive or negative. So, this part has no real roots. Since it's an \( x^8 \) polynomial, it must have 8 roots, and all of them must be imaginary. Adding these up, the original equation has 1 real root (which is 0) and 8 imaginary roots.

🎯 Exam Tip: When all coefficients of a polynomial with only even powers are positive, it will never have any real roots except for 0 if it has a common factor of x. This is because any real value (positive or negative) raised to an even power will be positive, making the sum of positive terms always positive, thus never equaling zero.