This MCQ module is based on: Linear Polynomials, Value & Zeros
TOPIC 5 OF 25
Linear Polynomials, Value & Zeros
🎓 Class 9
Mathematics
CBSE
Theory
Ch 2 — Introduction to Linear Polynomials
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Linear Polynomials, Value & Zeros
Targeting Class 9 level in Algebra, with Intermediate difficulty.
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2.2 Linear Polynomials — When the Variable's Power Is 1
An algebraic expression in one variable, where every term has the variable raised to at most the first power, is called a linear polynomial?.
Definition — Linear Polynomial
A polynomial in one variable \(x\) of the form
\[ p(x) = ax + b, \quad a \neq 0 \]
where \(a\) and \(b\) are real numbers, is called a linear polynomial. The number \(a\) is the leading coefficient (also the slope, as we'll see), and \(b\) is the constant term.
Example 4 — Side of a Square
Find the perimeter of a square whose side measures 1 cm, 1.5 cm, 2 cm, and 5.5 cm. Predict the perimeter for any side length \(x\) cm.
| Side \(x\) (cm) | 1 | 1.5 | 2 | 2.5 | 5.5 | \(x\) |
|---|---|---|---|---|---|---|
| Perimeter (cm) | 4 | 6 | 8 | 10 | 22 | \(4x\) |
Notice that whatever \(x\) is, the perimeter is exactly \(4x\). The expression \(p(x) = 4x\) is a linear polynomial.
Example 5 — Chess Tournament Fee
A chess club charges a joining fee of ₹200 plus ₹50 for every additional match played. The amount paid by a player after \(n\) matches:
| Matches \(n\) | 1 | 2 | 3 | 4 | 5 | ... | \(n\) |
|---|---|---|---|---|---|---|---|
| Amount (₹) | 250 | 300 | 350 | 400 | 450 | ... | \(200 + 50n\) |
Hence, if the number of matches played is \(n\), the total cost is given by the constant value of ₹200 plus ₹50 per match — a linear polynomial \(p(n) = 50n + 200\). For every additional match played, the amount payable increases by a constant ₹50.
Think and Reflect: If a player paid ₹750, how many matches did he play?
Solve \(50n + 200 = 750 \Rightarrow 50n = 550 \Rightarrow n = 11\). The player played 11 matches.
Linear Patterns — Key Examples
Examples 4 and 5 share a deep structural feature: the variable's power is 1, and changes happen at a constant rate. This is the hallmark of linear patterns.
Example 6 — Cycle Length and Width
Suppose two numbers \(x\) and \(y\) are such that the larger number minus twice the smaller equals 16, and the larger is 14 more than the smaller. Find the numbers.
Let larger \(= x\), smaller \(= y\). Equations: \(x - 2y = 16\) and \(x = y + 14\). Substituting: \((y + 14) - 2y = 16 \Rightarrow -y = 2 \Rightarrow y = -2\), so \(x = 12\). Numbers: \(\mathbf{12, -2}\).
Example 7 — Sum of Two Numbers
Find two numbers whose sum is 64 and whose difference is 12.
Let them be \(x\) and \(y\) with \(x > y\). \(x + y = 64\), \(x - y = 12\). Adding: \(2x = 76 \Rightarrow x = 38\). Then \(y = 26\). The numbers are 38 and 26.
2.3 Polynomials as Input–Output Machines
A polynomial like \(p(x) = 2x + 3\) is a function: feed it a number, get a number out. Some call this an input-output machine?.
Fig 2.3: A linear polynomial as an input-output machine
Value of a Polynomial
The value of a polynomial \(p(x)\) at \(x = c\) is the number obtained by substituting \(c\) in place of \(x\), denoted \(p(c)\). For \(p(x) = 2x + 3\): \(p(4) = 2(4) + 3 = 11\); \(p(0) = 3\); \(p(-1) = 1\).
Linear Decay — Example 8
The amount paid by a player decreases by ₹50 for each additional discount voucher used. If the base amount is ₹500 and \(x\) vouchers are used, the amount paid is
\(p(x) = 500 - 50x\) rupees.
This is also a linear polynomial — but the leading coefficient is negative. The amount decreases at a constant rate per voucher. We call this linear decay (vs. linear growth when the slope is positive).
Example 9 — Plant Height
A plant has height 1.75 ft and grows by 0.5 ft each month.
(i) Find the height after 7 months.
Height after \(n\) months = \(1.75 + 0.5n\). For \(n = 7\): \(1.75 + 3.5 = 5.25\) ft.
(ii) Make a table of values for \(n\) varying from 0 to 10.
n: 0,1,2,3,...,10 → h: 1.75, 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25, 5.75, 6.25, 6.75 ft.
(iii) Find an expression that relates \(h\) and \(n\).
\(h(n) = 0.5n + 1.75\) ft — a linear polynomial in \(n\) with slope 0.5 and intercept 1.75.
2.4 Zeros of a Polynomial
Definition — Zero of a Polynomial
A real number \(c\) is called a zero (or root) of the polynomial \(p(x)\) if \(p(c) = 0\). For a linear polynomial \(p(x) = ax + b\) (with \(a \neq 0\)) the unique zero is \(c = -\dfrac{b}{a}\).
Why exactly one zero?
Setting \(ax + b = 0\) and solving: \(x = -\dfrac{b}{a}\). Because \(a \neq 0\), this gives a single number — every linear polynomial has exactly one zero.
Worked Examples
Example 10: Find the zero of \(p(x) = 4x - 12\).
Set \(4x - 12 = 0 \Rightarrow x = 3\). Check: \(p(3) = 0\). ✓
Example 11: Find the zero of \(p(x) = -3x + 9\).
\(-3x + 9 = 0 \Rightarrow x = 3\). Note even when the leading coefficient is negative, the zero is computed the same way.
Example 12: Verify whether \(x = -2\) is a zero of \(p(x) = 2x + 4\).
\(p(-2) = 2(-2) + 4 = 0\). ✓ Yes, \(-2\) is a zero.
Activity: The Mystery Machine
L3 ApplyMaterials: Calculator, paper, partner
Predict: Can you guess your partner's secret linear polynomial after only 2 input-output pairs?
- Player A picks a secret linear polynomial \(p(x) = ax + b\) (with \(a, b\) integers between −5 and 5).
- Player B says any input \(x\); A returns \(p(x)\).
- After two trials, B must guess \(a\) and \(b\).
- Strategy: ask B \(p(0)\) — that gives \(b\). Then ask \(p(1)\) — that gives \(a + b\), so \(a\) follows.
- Switch roles; play 5 rounds.
Two well-chosen inputs always pin down a linear polynomial uniquely. This is the same principle behind "two points determine a line." More than 2 inputs would over-determine the line — and any redundant data must be consistent.
Competency-Based Questions
Scenario: A taxi company charges a fixed pickup fee of ₹50 plus ₹15 per kilometre. Let \(F(d)\) be the total fare for a ride of \(d\) km.
Q1. Write the linear polynomial for \(F(d)\) and compute the fare for an 8 km ride.
L3 Apply\(F(d) = 15d + 50\). For \(d=8\): \(F(8) = 120 + 50 = ₹170\).
Q2. A passenger has only ₹260. Analyse the maximum distance she can travel.
L4 Analyse\(15d + 50 \le 260 \Rightarrow 15d \le 210 \Rightarrow d \le 14\). Maximum = 14 km.
Q3. The driver claims that the "zero of \(F(d)\)" tells the kilometres at which the ride is free. Evaluate this claim.
L5 Evaluate\(F(d) = 0 \Rightarrow d = -50/15 \approx -3.33\) km, which is mathematically a zero but not physically meaningful — distance can't be negative. So the "zero" here exists algebraically but represents an impossible real-world state. The driver is wrong about a "free ride" interpretation.
Q4. Design a fare scheme \(F'(d) = ad + b\) that is cheaper than the existing one for short rides (d ≤ 5 km) but more expensive for long rides (d ≥ 15 km). Justify with calculations.
L6 CreateSample: \(F'(d) = 25d + 20\). At \(d=5\): \(F = 125, F' = 145\) — wait, that's higher. Try \(F'(d) = 10d + 30\). At d=5: F'=80 (vs 125 ✓), at d=15: F'=180 (vs 275... still cheaper). Adjusted: \(F'(d) = 22d + 0\). At d=5: F'=110 (vs 125 ✓), at d=15: F'=330 (vs 275 ✗ ✓ more expensive). Many valid designs.
Assertion–Reason Questions
A: The polynomial \(p(x) = 5x - 10\) has exactly one zero.
R: Every linear polynomial \(ax + b\) with \(a \neq 0\) has exactly one zero, namely \(-b/a\).
R: Every linear polynomial \(ax + b\) with \(a \neq 0\) has exactly one zero, namely \(-b/a\).
Answer: (a) — Zero of \(5x-10\) is \(x = 2\). R explains A.
A: The polynomial \(p(x) = 7\) has no zero.
R: A constant non-zero polynomial cannot equal zero for any value of \(x\).
R: A constant non-zero polynomial cannot equal zero for any value of \(x\).
Answer: (a) — \(p(x) = 7\) is constant, never 0. R is precisely the reason.
A: If \(p(x) = 2x + 6\), then \(p(-3) = 0\).
R: The zero of \(p(x) = 2x + 6\) is \(x = 3\).
R: The zero of \(p(x) = 2x + 6\) is \(x = 3\).
Answer: (c) — A is correct: \(2(-3)+6=0\). R is wrong: zero is \(-3\), not 3.
Frequently Asked Questions
What is a linear polynomial in Class 9 Maths?
A linear polynomial in one variable is p(x) = ax + b where a and b are real numbers and a is not zero. The highest power of x is one, so its graph is a straight line.
How do you find the value of a linear polynomial at x = k?
Substitute x = k into the polynomial. For p(x) = ax + b, the value at k is p(k) = a*k + b. For example, if p(x) = 2x + 3 then p(4) = 2*4 + 3 = 11.
What is the zero of a linear polynomial?
The zero (or root) of a polynomial is the value of x that makes p(x) = 0. For p(x) = ax + b, setting ax + b = 0 gives the unique zero x = -b/a, provided a is not zero.
How many zeros does a linear polynomial have?
A linear polynomial p(x) = ax + b with a not equal to 0 has exactly one zero, given by x = -b/a. This corresponds to the single point where its straight-line graph crosses the x-axis.
What is the difference between a linear polynomial and a linear equation?
A linear polynomial p(x) = ax + b is an expression. Setting it equal to a value, such as ax + b = 0 or ax + b = c, turns it into a linear equation that we solve for x. Every linear polynomial gives rise to a linear equation when equated to zero.
Can the coefficient a in p(x) = ax + b be zero?
No. If a = 0 the expression becomes the constant b and the variable disappears, so it is no longer linear. The condition a is not equal to 0 is essential for ax + b to be a linear polynomial.
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