Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra Exercise 3.2

Question 1. Find the value of the polynomial \( f(y) = 6y - 3y^2 + 3 \) at
(i) \( y = 1 \)
(ii) \( y = -1 \)
(iii) \( y = 0 \)
Answer:
(i) When \( y = 1 \):
\( f(y) = 6y - 3y^2 + 3 \)
\( f(1) = 6(1) - 3(1)^2 + 3 \)
\( = 6 - 3 + 3 \)
\( = 6 \)

(ii) When \( y = -1 \):
\( f(y) = 6y - 3y^2 + 3 \)
\( f(-1) = 6(-1) - 3(-1)^2 + 3 \)
\( = -6 - 3 + 3 \)
\( = -6 \)

(iii) When \( y = 0 \):
\( f(y) = 6y - 3y^2 + 3 \)
\( f(0) = 6(0) - 3(0)^2 + 3 \)
\( = 0 - 0 + 3 \)
\( = 3 \)
The value of a polynomial changes based on the value we substitute for the variable. This is why polynomials are useful in modelling situations where quantities change.
In simple words: We put the given 'y' value into the polynomial equation. Then, we do the math step by step to find the final number for each case.

🎯 Exam Tip: Always follow the order of operations (PEMDAS/BODMAS) carefully, especially when dealing with negative numbers and exponents, to avoid errors.

Question 2. If \( p(x) = x^2 - 2\sqrt{2}x + 1 \), find \( p(2\sqrt{2}) \).
Answer:
Given the polynomial \( p(x) = x^2 - 2\sqrt{2}x + 1 \).
We need to find the value of \( p(x) \) when \( x = 2\sqrt{2} \).
Substitute \( x = 2\sqrt{2} \) into the polynomial:
\( p(2\sqrt{2}) = (2\sqrt{2})^2 - 2\sqrt{2}(2\sqrt{2}) + 1 \)
\( = (4 \times 2) - (4 \times 2) + 1 \)
\( = 8 - 8 + 1 \)
\( = 0 + 1 \)
\( = 1 \)
Remember that \( (\sqrt{a})^2 = a \), which simplifies calculations involving square roots. So, \( (2\sqrt{2})^2 = 2^2 \times (\sqrt{2})^2 = 4 \times 2 = 8 \).
In simple words: Replace 'x' in the equation with \( 2\sqrt{2} \). Calculate each part carefully, especially the square roots and squares. The final answer is 1.

🎯 Exam Tip: When substituting values with square roots, square both the coefficient and the square root part, e.g., \( (2\sqrt{2})^2 = 2^2 \times (\sqrt{2})^2 = 4 \times 2 = 8 \).

Question 3. Find the zeros of the polynomial in each of the following.
(i) \( P(x) = x - 3 \)
(ii) \( p(x) = 2x + 5 \)
(iii) \( q(y) = 2y - 3 \)
(iv) \( f(z) = 8z \)
(v) \( p(x) = ax \) when \( a \neq 0 \)
(vi) \( h(x) = ax + b \), where \( a \neq 0 \) and \( a, b \in R \)
Answer:
(i) To find the zero of the polynomial \( P(x) = x - 3 \), we set \( P(x) = 0 \):
\( x - 3 = 0 \)
\( \implies x = 3 \)
Therefore, \( 3 \) is the zero of the polynomial \( P(x) \). A zero of a polynomial is the value of the variable that makes the polynomial equal to zero.
In simple words: To find the zero, make the polynomial equal to zero and solve for 'x'. For \( x-3=0 \), 'x' must be 3.

(ii) To find the zero of the polynomial \( p(x) = 2x + 5 \), we set \( p(x) = 0 \):
\( 2x + 5 = 0 \)
\( \implies 2x = -5 \)
\( \implies x = -\frac{5}{2} \)
So, \( -\frac{5}{2} \) is the zero of the polynomial \( p(x) \). Finding zeros helps us understand where the polynomial's graph crosses the x-axis.
In simple words: Set \( 2x+5 \) to zero. Move 5 to the other side to get \( 2x = -5 \). Then divide by 2, so 'x' is \( -\frac{5}{2} \).

(iii) To find the zero of the polynomial \( q(y) = 2y - 3 \), we set \( q(y) = 0 \):
\( 2y - 3 = 0 \)
\( \implies 2y = 3 \)
\( \implies y = \frac{3}{2} \)
Thus, \( \frac{3}{2} \) is the zero of the polynomial \( q(y) \). The variable can be 'x', 'y', or 'z', but the process to find the zero remains the same.
In simple words: Make \( 2y-3 \) equal to zero. This means \( 2y = 3 \), so 'y' is \( \frac{3}{2} \).

(iv) To find the zero of the polynomial \( f(z) = 8z \), we set \( f(z) = 0 \):
\( 8z = 0 \)
\( \implies z = \frac{0}{8} \)
\( \implies z = 0 \)
Therefore, \( 0 \) is the zero of the polynomial \( f(z) \). When a polynomial is just a constant multiplied by a variable, its zero is always 0.
In simple words: Set \( 8z \) to zero. If 8 times 'z' is 0, then 'z' must be 0.

(v) To find the zero of the polynomial \( p(x) = ax \) (where \( a \neq 0 \)), we set \( p(x) = 0 \):
\( ax = 0 \)
Since \( a \neq 0 \), we must have:
\( \implies x = \frac{0}{a} \)
\( \implies x = 0 \)
So, \( 0 \) is the zero of the polynomial \( p(x) \). This highlights that multiplying any non-zero number by zero always results in zero.
In simple words: If 'a' times 'x' is zero, and 'a' is not zero, then 'x' has to be zero.

(vi) To find the zero of the polynomial \( h(x) = ax + b \) (where \( a \neq 0 \)), we set \( h(x) = 0 \):
\( ax + b = 0 \)
\( \implies ax = -b \)
\( \implies x = -\frac{b}{a} \)
Therefore, \( -\frac{b}{a} \) is the zero of the polynomial \( h(x) \). This is a general form for the zero of any linear polynomial.
In simple words: To make \( ax+b \) equal to zero, we first move 'b' to the other side (making it \( -b \)). Then we divide by 'a', so 'x' is \( -\frac{b}{a} \).

🎯 Exam Tip: To find the zero of a polynomial, always set the polynomial expression equal to zero and solve for the variable. This is a fundamental concept in algebra.

Question 4. Find the roots of the polynomial equations.
(i) \( 5x - 6 = 0 \)
(ii) \( x + 3 = 0 \)
(iii) \( 10x + 9 = 0 \)
(iv) \( 9x - 4 = 0 \)
Answer:
(i) To find the root of the equation \( 5x - 6 = 0 \), we solve for \( x \):
\( 5x - 6 = 0 \)
\( \implies 5x = 6 \)
\( \implies x = \frac{6}{5} \)
So, \( \frac{6}{5} \) is the root of the polynomial equation. Roots are essentially the same as zeros for polynomial equations.
In simple words: To find 'x', first add 6 to both sides, so \( 5x=6 \). Then divide by 5, making \( x = \frac{6}{5} \).

(ii) To find the root of the equation \( x + 3 = 0 \), we solve for \( x \):
\( x + 3 = 0 \)
\( \implies x = -3 \)
Therefore, \( -3 \) is the root of the polynomial equation. It's the value that perfectly balances the equation.
In simple words: To find 'x', subtract 3 from both sides, which gives \( x = -3 \).

(iii) To find the root of the equation \( 10x + 9 = 0 \), we solve for \( x \):
\( 10x + 9 = 0 \)
\( \implies 10x = -9 \)
\( \implies x = -\frac{9}{10} \)
Hence, \( -\frac{9}{10} \) is the root of the polynomial equation. Remember to change the sign when moving terms across the equals sign.
In simple words: Subtract 9 from both sides, so \( 10x = -9 \). Then divide by 10, meaning \( x = -\frac{9}{10} \).

(iv) To find the root of the equation \( 9x - 4 = 0 \), we solve for \( x \):
\( 9x - 4 = 0 \)
\( \implies 9x = 4 \)
\( \implies x = \frac{4}{9} \)
Thus, \( \frac{4}{9} \) is the root of the polynomial equation. Always ensure your final answer is in its simplest fraction form.
In simple words: Add 4 to both sides, so \( 9x = 4 \). Then divide by 9, so \( x = \frac{4}{9} \).

🎯 Exam Tip: The terms "roots" of a polynomial equation and "zeros" of a polynomial are often used interchangeably when the polynomial is set to zero. Both refer to the values that make the expression equal to zero.

Question 5. Verify whether the following are zeros of the polynomial, indicated against them, or not.
(i) \( p(x) = 2x - 1 \), \( x = \frac{1}{2} \)
(ii) \( p(x) = x^3 - 1 \), \( x = 1 \)
(iii) \( p(x) = ax + b \), \( x = -\frac{b}{a} \)
(iv) \( p(x) = (x + 3) (x - 4) \); \( x = -3, x = 4 \)
Answer:
(i) Given polynomial \( p(x) = 2x - 1 \) and value \( x = \frac{1}{2} \).
Substitute \( x = \frac{1}{2} \) into the polynomial:
\( p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right) - 1 \)
\( = 1 - 1 \)
\( = 0 \)
Since \( p\left(\frac{1}{2}\right) = 0 \), \( \frac{1}{2} \) is indeed a zero of the polynomial \( p(x) \). This confirms the definition of a zero.
In simple words: Put \( \frac{1}{2} \) in place of 'x'. If the answer is zero, then \( \frac{1}{2} \) is a zero of the polynomial. Here, it is.

(ii) Given polynomial \( p(x) = x^3 - 1 \) and value \( x = 1 \).
Substitute \( x = 1 \) into the polynomial:
\( p(1) = (1)^3 - 1 \)
\( = 1 - 1 \)
\( = 0 \)
Since \( p(1) = 0 \), \( 1 \) is a zero of the polynomial \( p(x) \). This is a simple case often used to demonstrate the concept of a zero.
In simple words: Put 1 in place of 'x'. \( 1^3 \) is 1, and \( 1-1 \) is 0. So, 1 is a zero.

(iii) Given polynomial \( p(x) = ax + b \) and value \( x = -\frac{b}{a} \).
Substitute \( x = -\frac{b}{a} \) into the polynomial:
\( p\left(-\frac{b}{a}\right) = a\left(-\frac{b}{a}\right) + b \)
\( = -b + b \)
\( = 0 \)
Since \( p\left(-\frac{b}{a}\right) = 0 \), \( -\frac{b}{a} \) is a zero of the polynomial \( p(x) \). This result holds true for any linear polynomial in general form.
In simple words: Put \( -\frac{b}{a} \) in place of 'x'. When you multiply 'a' by \( -\frac{b}{a} \), you get \( -b \). Then \( -b+b \) is 0. So it is a zero.

(iv) Given polynomial \( p(x) = (x + 3)(x - 4) \) and values \( x = -3, x = 4 \).
First, check for \( x = -3 \):
\( p(-3) = (-3 + 3)(-3 - 4) \)
\( = (0)(-7) \)
\( = 0 \)
Next, check for \( x = 4 \):
\( p(4) = (4 + 3)(4 - 4) \)
\( = (7)(0) \)
\( = 0 \)
Since both \( p(-3) = 0 \) and \( p(4) = 0 \), both \( -3 \) and \( 4 \) are zeros of the polynomial \( p(x) \). This demonstrates that a polynomial can have multiple zeros.
In simple words: First, put \( -3 \) for 'x'. You get \( (0) \times (-7) = 0 \). Then, put 4 for 'x'. You get \( (7) \times (0) = 0 \). Since both give zero, both \( -3 \) and 4 are zeros.

🎯 Exam Tip: To verify if a given value is a zero of a polynomial, substitute the value into the polynomial expression. If the result is zero, then the value is indeed a zero.

Question 6. Find the number of zeros of the following polynomials represented by their graphs.
Answer:
The number of zeros of a polynomial represented by a graph is equal to the number of times its curve intersects or touches the x-axis. Here are the counts for each graph:
(i) Number of zeros = 2 (The curve intersects the x-axis at two distinct points).
(ii) Number of zeros = 3 (The curve intersects the x-axis at three distinct points).
(iii) Number of zeros = 0 (The curve does not intersect the x-axis at all).
(iv) Number of zeros = 1 (The curve intersects the x-axis at the origin, which is one point).
(v) Number of zeros = 1 (The curve intersects the x-axis at one distinct point).
Each point where the graph crosses the x-axis corresponds to a real zero of the polynomial.
In simple words: Just count how many times the graph crosses the horizontal line (x-axis). That number tells you how many zeros the polynomial has.

🎯 Exam Tip: Remember that a zero corresponds to an x-intercept. Carefully count how many times the graph crosses or touches the x-axis to determine the number of real zeros.