Building Polynomials with Specified Zeros Step 1: Learning Polynomial Building - Provided zeros: -3 and 5 - Simple polynomial form: (x + 3)(x - 5) - Expanding: x² - 2x - 15 Step 2: Freedom Degree Polynomials may be formed by multiplying the simple form by any non-zero constant. Possible Polynomials:Read more
General Form:
For any non-zero constant k:
k(x² – 2x – 15)
Mathematical Insight:
– The k is the parameter for infinite scaling of the polynomial
– Each scale produces a different polynomial with the same zeros
– These are actually the same polynomial to a constant factor
Step 3: Counting Polynomials
– We can make polynomials with these zeros by multiplying the fundamental form
by any real number
– This produces an infinite number of polynomials
Conclusion
There are MORE THAN 3 polynomials with zeros -3 and 5.
Building a Quadratic Polynomial with Given Zeros Step 1: Defining Zeros - Given zeros: 4 and -3 - The polynomial shall thus be in the form: (x - 4)(x + 3) Step 2: Algebraic Expansion (x - 4)(x + 3) = x² + 3x - 4x - 12 = x² - x - 12 Step 3: Checking Zero Properties Let's test if the zeros hold: - WheRead more
Building a Quadratic Polynomial with Given Zeros
Step 1: Defining Zeros
– Given zeros: 4 and -3
– The polynomial shall thus be in the form: (x – 4)(x + 3)
Step 3: Checking Zero Properties
Let’s test if the zeros hold:
– When x = 4:
4² – 4 – 12 = 16 – 4 – 12 = 0 OK
– When x = -3:
(-3)² – (-3) – 12 = 9 + 3 – 12 = 0 OK
Step 4: Coefficient Analysis
– Coefficient of x²: 1
– Coefficient of x: -1
– Constant term: -12
Mathematical Insights:
– The zeros of a quadratic define its form
– The general form illustrates how the zeros are connected to the coefficients of the polynomial
– Vieta’s formulas support the connection between coefficients and zeros
Conclusion:
The quadratic polynomial whose zeros are 4 and -3 is x² – x – 12.
Solving for the Unknown Zero of a Quadratic Polynomial Step 1: Given Information - Polynomial: 2x² + 7x + 3 - One known zero: -3/2 Step 2: Vieta's Formulas for Quadratic Polynomials Let the two zeros be p and q: - p + q = -b/a - p * q = c/a Where in 2x² + 7x + 3: - a = 2 - b = 7 - c = 3 Step 3: UsinRead more
Solving for the Unknown Zero of a Quadratic Polynomial
Step 1: Given Information
– Polynomial: 2x² + 7x + 3
– One known zero: -3/2
Step 2: Vieta’s Formulas for Quadratic Polynomials
Let the two zeros be p and q:
– p + q = -b/a
– p * q = c/a
Where in 2x² + 7x + 3:
– a = 2
– b = 7
– c = 3
Step 3: Using the Known Zero
If p = -3/2, then we need to find q
Sum of zeros formula:
p + q = -b/a
-3/2 + q = -7/2
Step 4: Solving for the Unknown Zero
q = -7/2 – (-3/2)
= -7/2 + 3/2
= -4/2
= -2
Verification:
– First zero: p = -3/2
– Second zero: q = -2
– Check: (-3/2) + (-2) = -7/2 ✓
– Check: (-3/2) * (-2) = 3/a ✓
Mathematical Insight:
Vieta’s formulas provide a powerful method to find
polynomial zeros without fully solving the equation.
Conclusion:
The other zero of the polynomial is -2 (or 1 in the given options).
Determining the Remainder of Polynomial Division Step 1: Remainder Theorem Basics - The remainder theorem says that when a polynomial f(x) is divided by (x - a), the remainder is f(a) - This implies we can determine the remainder by evaluating the polynomial at x = 2 Step 2: Evaluating a PolynomialRead more
Determining the Remainder of Polynomial Division
Step 1: Remainder Theorem Basics
– The remainder theorem says that when a polynomial f(x) is divided by (x – a),
the remainder is f(a)
– This implies we can determine the remainder by evaluating the polynomial at x = 2
Step 2: Evaluating a Polynomial
Polynomial: f(x) = x³ – 6x² + 11x – 6
Substituting x = 2:
Mathematical Insight:
– By simply putting the root of the divisor (2) in the polynomial
– We can easily find the remainder without long division
– It is a very strong method that makes polynomial remainder computation easier
Step 3: Confirmation
– The remainder is 0
– It indicates (x – 2) evenly divides the polynomial
– There is no remaind value when divided
Conclusion:
The remainder upon division of x³ – 6x² + 11x – 6 by (x – 2) is zero.
Constructing a Quadratic Polynomial with Specific Zero Properties Step 1: Understanding Vieta's Formulas For a quadratic polynomial ax² + bx + c with zeros p and q: - Sum of zeros: p + q = -b/a - Product of zeros: p * q = c/a Given Conditions: - Sum of zeros = -4 - Product of zeros = 3 Step 2: AnalyRead more
Constructing a Quadratic Polynomial with Specific Zero Properties
Step 1: Understanding Vieta’s Formulas
For a quadratic polynomial ax² + bx + c with zeros p and q:
– Sum of zeros: p + q = -b/a
– Product of zeros: p * q = c/a
Given Conditions:
– Sum of zeros = -4
– Product of zeros = 3
Step 2: Analyzing the Coefficients
Let’s consider a standard quadratic form: x² + 4x + c
Checking Sum of Zeros:
– p + q = -4
– This means the coefficient of x must be -4
Checking Product of Zeros:
– p * q = 3
– This means the constant term must be 3
Checking Conditions:
– Sum of zeros: 3 + 1 = 4 ✓
– Product of zeros: 3 * 1 = 3 ✓
Key Insights:
– Vieta’s formulas provide a powerful way to relate
zeros to polynomial coefficients
– We can construct polynomials by understanding
the relationships between zeros and coefficients
Conclusion:
The polynomial that satisfies the given conditions is x² – 4x + 3.
Understanding Zeros of a Cubic Polynomial Mathematical Background A cubic polynomial is of the general form: ax³ + bx² + cx + d, where a ≠ 0 Fundamental Theorem of Algebra - Any polynomial has exactly as many zeros as its degree - These zeros can be real or complex numbers - In a cubic polynomial, tRead more
Understanding Zeros of a Cubic Polynomial
Mathematical Background
A cubic polynomial is of the general form:
ax³ + bx² + cx + d, where a ≠ 0
Fundamental Theorem of Algebra
– Any polynomial has exactly as many zeros as its degree
– These zeros can be real or complex numbers
– In a cubic polynomial, these zeros are referred to as “roots”
Mere Zero Analysis
– A cubic polynomial ALWAYS has 3 zeros
– These zeros can include:
– 3 real zeros in different positions
– 1 real zero and 2 complex conjugate zeros
– 1 repeated real zero that occurs twice
– One real zero that occurs thrice
Mathematical Proof Highlights
– The Fundamental Theorem of Algebra assures 3 zeros
– Complex numbers make all polynomials fully factorable
– Mathematically denoted as:
ax³ + bx² + cx + d = a(x – r₁)(x – r₂)(x – r₃)
Where r₁, r₂, r₃ are the three zeros
Key Insight
The degree of the polynomial always decides the number of zeros,
never mind the actual type of those zeros.
Conclusion:
A cubic polynomial ALWAYS has 3 zeros.
Finding the Second Zero of a Quadratic Polynomial Step 1: Understanding the Given Information - Polynomial: x² – 7x + 10 - One known zero: 5 Step 2: Verification of the Known Zero Let's first verify that 5 is indeed a zero: 5² – 7(5) + 10 = 25 – 35 + 10 = 0 Step 3: Using Vieta's Formulas In a quadraRead more
Finding the Second Zero of a Quadratic Polynomial
Step 1: Understanding the Given Information
– Polynomial: x² – 7x + 10
– One known zero: 5
Step 2: Verification of the Known Zero
Let’s first verify that 5 is indeed a zero:
5² – 7(5) + 10 = 25 – 35 + 10 = 0
Step 3: Using Vieta’s Formulas
In a quadratic polynomial ax² + bx + c, if p and q are zeros:
– Sum of zeros: p + q = -b/a
– Product of zeros: p * q = c/a
For x² – 7x + 10:
– a = 1
– b = -7
– c = 10
Step 4: Finding the Second Zero
We are aware that one zero is 5, therefore let’s use the variable x to represent the second zero.
Mathematical Insight:
Vieta’s formulas offer a beautiful method of determining polynomial zeros
without resorting to complicated solving methods. They show the profound
connection between a polynomial’s coefficients and its roots.
Conclusion:
The other zero of the polynomial is 2.
Step 1: Definition of Standard Form Standard form of a linear equation is represented as: Ax + By = C Where: - A, B, and C are constants - A, B, and C are integers - A and B are not both zero - A ≥ 0 (if A = 0, then B must be positive) Step 2: Given Equation Analysis Original equation: x + 2y = 3 StRead more
Step 1: Definition of Standard Form
Standard form of a linear equation is represented as:
Ax + By = C
Where:
– A, B, and C are constants
– A, B, and C are integers
– A and B are not both zero
– A ≥ 0 (if A = 0, then B must be positive)
Step 2: Given Equation Analysis
Original equation: x + 2y = 3
Step 3: Transformation to Standard Form
– The equation is already very close to standard form
– To make it exactly standard form, we need to move all terms to one side
– Rearrange to: x + 2y – 3 = 0
Verification:
– Coefficients of x: 1
– Coefficients of y: 2
– Constant term: -3
– All parts meet standard form requirements
Step 4: Why Other Options Are Incorrect
– “2y + x = 3” ≠ standard form (terms not on one side)
– “3 – x = 2y” ≠ standard form (rearranged incorrectly)
Conclusion:
The correct standard form is x + 2y – 3 = 0
Step 1: Definition of Inconsistent Equations - Inconsistent equations are linear equations that have NO SOLUTION - Graphically, this means the lines representing these equations NEVER intersect Step 2: Graphical Interpretation Inconsistent equations always result in PARALLEL LINES - These lines haveRead more
Step 1: Definition of Inconsistent Equations
– Inconsistent equations are linear equations that have NO SOLUTION
– Graphically, this means the lines representing these equations NEVER intersect
Step 2: Graphical Interpretation
Inconsistent equations always result in PARALLEL LINES
– These lines have the same slope but different y-intercepts
– They run alongside each other, maintaining a constant distance
– No point exists where these lines cross
Step 3: Algebraic Characteristics
Example of Inconsistent Equations:
– Equation ₁: 2x + y = 4
– Equation ₂: 2x + y = 6
Observe:
– Same coefficient for x (2)
– Same coefficient for y (1)
– Different constant terms (4 and 6)
– This guarantees the lines will be parallel
Step 4: Geometric Visualization
– Imagine two identical lines shifted vertically
– They have the same “direction” but never touch
– No matter how far you extend them, they remain parallel
Conclusion:
For a pair of linear equations to be inconsistent, their graphs MUST be PARALLEL LINES.
Step 1: Equation Analysis Equation ₁: 2x + 3y = 5 Equation ₂: 4x + 6y = 10 Step 2: Proportionality Check - Coefficient of x in Equation ₁: 2 - Coefficient of x in Equation ₂: 4 - Ratio of x coefficients: 4 ÷ 2 = 2 - Coefficient of y in Equation ₁: 3 - Coefficient of y in Equation ₂: 6 - Ratio of y cRead more
Step 2: Proportionality Check
– Coefficient of x in Equation ₁: 2
– Coefficient of x in Equation ₂: 4
– Ratio of x coefficients: 4 ÷ 2 = 2
– Coefficient of y in Equation ₁: 3
– Coefficient of y in Equation ₂: 6
– Ratio of y coefficients: 6 ÷ 3 = 2
Step 3: Constant Term Verification
– Equation ₁ constant: 5
– Equation ₂ constant: 10
– Ratio of constants: 10 ÷ 5 = 2
Step 4: Algebraic Manipulation
Divide Equation ₂ by 2:
2x + 3y = 5 (Identical to Equation ₁)
Step 5: Interpretation
– The equations represent EXACTLY THE SAME LINE
– This means the system has INFINITELY MANY SOLUTIONS
– Every point on this line satisfies both equations
Mathematical Insight:
When two linear equations represent identical lines,
they have an infinite number of solution points that
perfectly overlap each other.
Conclusion:
The system has INFINITELY MANY SOLUTIONS.
The number of polynomials having zeros -3 and 5 is
Building Polynomials with Specified Zeros Step 1: Learning Polynomial Building - Provided zeros: -3 and 5 - Simple polynomial form: (x + 3)(x - 5) - Expanding: x² - 2x - 15 Step 2: Freedom Degree Polynomials may be formed by multiplying the simple form by any non-zero constant. Possible Polynomials:Read more
Building Polynomials with Specified Zeros
Step 1: Learning Polynomial Building
– Provided zeros: -3 and 5
– Simple polynomial form: (x + 3)(x – 5)
– Expanding: x² – 2x – 15
Step 2: Freedom Degree
Polynomials may be formed by multiplying the simple form by any non-zero constant.
Possible Polynomials:
1. x² – 2x – 15
2. 2x² – 4x – 30
3. 3x² – 6x – 45
General Form:
For any non-zero constant k:
k(x² – 2x – 15)
Mathematical Insight:
– The k is the parameter for infinite scaling of the polynomial
– Each scale produces a different polynomial with the same zeros
– These are actually the same polynomial to a constant factor
Step 3: Counting Polynomials
– We can make polynomials with these zeros by multiplying the fundamental form
by any real number
– This produces an infinite number of polynomials
Conclusion
There are MORE THAN 3 polynomials with zeros -3 and 5.
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The quadratic polynomial whose zeros are 4 and -3 is:
Building a Quadratic Polynomial with Given Zeros Step 1: Defining Zeros - Given zeros: 4 and -3 - The polynomial shall thus be in the form: (x - 4)(x + 3) Step 2: Algebraic Expansion (x - 4)(x + 3) = x² + 3x - 4x - 12 = x² - x - 12 Step 3: Checking Zero Properties Let's test if the zeros hold: - WheRead more
Building a Quadratic Polynomial with Given Zeros
Step 1: Defining Zeros
– Given zeros: 4 and -3
– The polynomial shall thus be in the form: (x – 4)(x + 3)
Step 2: Algebraic Expansion
(x – 4)(x + 3) = x² + 3x – 4x – 12
= x² – x – 12
Step 3: Checking Zero Properties
Let’s test if the zeros hold:
– When x = 4:
4² – 4 – 12 = 16 – 4 – 12 = 0 OK
– When x = -3:
(-3)² – (-3) – 12 = 9 + 3 – 12 = 0 OK
Step 4: Coefficient Analysis
– Coefficient of x²: 1
– Coefficient of x: -1
– Constant term: -12
Mathematical Insights:
– The zeros of a quadratic define its form
– The general form illustrates how the zeros are connected to the coefficients of the polynomial
– Vieta’s formulas support the connection between coefficients and zeros
Conclusion:
The quadratic polynomial whose zeros are 4 and -3 is x² – x – 12.
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If one of the zeros of the polynomial 2x² + 7x + 3 is -3/2, then the other zero is:
Solving for the Unknown Zero of a Quadratic Polynomial Step 1: Given Information - Polynomial: 2x² + 7x + 3 - One known zero: -3/2 Step 2: Vieta's Formulas for Quadratic Polynomials Let the two zeros be p and q: - p + q = -b/a - p * q = c/a Where in 2x² + 7x + 3: - a = 2 - b = 7 - c = 3 Step 3: UsinRead more
Solving for the Unknown Zero of a Quadratic Polynomial
Step 1: Given Information
– Polynomial: 2x² + 7x + 3
– One known zero: -3/2
Step 2: Vieta’s Formulas for Quadratic Polynomials
Let the two zeros be p and q:
– p + q = -b/a
– p * q = c/a
Where in 2x² + 7x + 3:
– a = 2
– b = 7
– c = 3
Step 3: Using the Known Zero
If p = -3/2, then we need to find q
Sum of zeros formula:
p + q = -b/a
-3/2 + q = -7/2
Step 4: Solving for the Unknown Zero
q = -7/2 – (-3/2)
= -7/2 + 3/2
= -4/2
= -2
Verification:
– First zero: p = -3/2
– Second zero: q = -2
– Check: (-3/2) + (-2) = -7/2 ✓
– Check: (-3/2) * (-2) = 3/a ✓
Mathematical Insight:
Vieta’s formulas provide a powerful method to find
polynomial zeros without fully solving the equation.
Conclusion:
The other zero of the polynomial is -2 (or 1 in the given options).
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
If the polynomial x³ – 6x² + 11x – 6 is divided by (x – 2), then the remainder is:
Determining the Remainder of Polynomial Division Step 1: Remainder Theorem Basics - The remainder theorem says that when a polynomial f(x) is divided by (x - a), the remainder is f(a) - This implies we can determine the remainder by evaluating the polynomial at x = 2 Step 2: Evaluating a PolynomialRead more
Determining the Remainder of Polynomial Division
Step 1: Remainder Theorem Basics
– The remainder theorem says that when a polynomial f(x) is divided by (x – a),
the remainder is f(a)
– This implies we can determine the remainder by evaluating the polynomial at x = 2
Step 2: Evaluating a Polynomial
Polynomial: f(x) = x³ – 6x² + 11x – 6
Substituting x = 2:
f(2) = 2³ – 6(2)² + 11(2) – 6
= 8 – 6(4) + 22 – 6
= 8 – 24 + 22 – 6
= 0
Mathematical Insight:
– By simply putting the root of the divisor (2) in the polynomial
– We can easily find the remainder without long division
– It is a very strong method that makes polynomial remainder computation easier
Step 3: Confirmation
– The remainder is 0
– It indicates (x – 2) evenly divides the polynomial
– There is no remaind value when divided
Conclusion:
The remainder upon division of x³ – 6x² + 11x – 6 by (x – 2) is zero.
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See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
If the sum of the zeros of the quadratic polynomial ax² + bx + c is -4 and the product of the zeros is 3, then the polynomial is:
Constructing a Quadratic Polynomial with Specific Zero Properties Step 1: Understanding Vieta's Formulas For a quadratic polynomial ax² + bx + c with zeros p and q: - Sum of zeros: p + q = -b/a - Product of zeros: p * q = c/a Given Conditions: - Sum of zeros = -4 - Product of zeros = 3 Step 2: AnalyRead more
Constructing a Quadratic Polynomial with Specific Zero Properties
Step 1: Understanding Vieta’s Formulas
For a quadratic polynomial ax² + bx + c with zeros p and q:
– Sum of zeros: p + q = -b/a
– Product of zeros: p * q = c/a
Given Conditions:
– Sum of zeros = -4
– Product of zeros = 3
Step 2: Analyzing the Coefficients
Let’s consider a standard quadratic form: x² + 4x + c
Checking Sum of Zeros:
– p + q = -4
– This means the coefficient of x must be -4
Checking Product of Zeros:
– p * q = 3
– This means the constant term must be 3
Step 3: Verification
The polynomial becomes: x² – 4x + 3
Mathematical Verification:
Let’s find the zeros using the quadratic formula:
x = [4 ± √(16 – 4(1)(3))] / 2(1)
= [4 ± √(16 – 12)] / 2
= [4 ± √4] / 2
= [4 ± 2] / 2
Zeros are:
– p = (4 + 2)/2 = 3
– q = (4 – 2)/2 = 1
Checking Conditions:
– Sum of zeros: 3 + 1 = 4 ✓
– Product of zeros: 3 * 1 = 3 ✓
Key Insights:
– Vieta’s formulas provide a powerful way to relate
zeros to polynomial coefficients
– We can construct polynomials by understanding
the relationships between zeros and coefficients
Conclusion:
The polynomial that satisfies the given conditions is x² – 4x + 3.
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The number of zeros in a cubic polynomial is always:
Understanding Zeros of a Cubic Polynomial Mathematical Background A cubic polynomial is of the general form: ax³ + bx² + cx + d, where a ≠ 0 Fundamental Theorem of Algebra - Any polynomial has exactly as many zeros as its degree - These zeros can be real or complex numbers - In a cubic polynomial, tRead more
Understanding Zeros of a Cubic Polynomial
Mathematical Background
A cubic polynomial is of the general form:
ax³ + bx² + cx + d, where a ≠ 0
Fundamental Theorem of Algebra
– Any polynomial has exactly as many zeros as its degree
– These zeros can be real or complex numbers
– In a cubic polynomial, these zeros are referred to as “roots”
Mere Zero Analysis
– A cubic polynomial ALWAYS has 3 zeros
– These zeros can include:
– 3 real zeros in different positions
– 1 real zero and 2 complex conjugate zeros
– 1 repeated real zero that occurs twice
– One real zero that occurs thrice
Mathematical Proof Highlights
– The Fundamental Theorem of Algebra assures 3 zeros
– Complex numbers make all polynomials fully factorable
– Mathematically denoted as:
ax³ + bx² + cx + d = a(x – r₁)(x – r₂)(x – r₃)
Where r₁, r₂, r₃ are the three zeros
Key Insight
The degree of the polynomial always decides the number of zeros,
never mind the actual type of those zeros.
Conclusion:
A cubic polynomial ALWAYS has 3 zeros.
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
If one zero of the polynomial x² – 7x + 10 is 5, then the other zero is:
Finding the Second Zero of a Quadratic Polynomial Step 1: Understanding the Given Information - Polynomial: x² – 7x + 10 - One known zero: 5 Step 2: Verification of the Known Zero Let's first verify that 5 is indeed a zero: 5² – 7(5) + 10 = 25 – 35 + 10 = 0 Step 3: Using Vieta's Formulas In a quadraRead more
Finding the Second Zero of a Quadratic Polynomial
Step 1: Understanding the Given Information
– Polynomial: x² – 7x + 10
– One known zero: 5
Step 2: Verification of the Known Zero
Let’s first verify that 5 is indeed a zero:
5² – 7(5) + 10 = 25 – 35 + 10 = 0
Step 3: Using Vieta’s Formulas
In a quadratic polynomial ax² + bx + c, if p and q are zeros:
– Sum of zeros: p + q = -b/a
– Product of zeros: p * q = c/a
For x² – 7x + 10:
– a = 1
– b = -7
– c = 10
Step 4: Finding the Second Zero
We are aware that one zero is 5, therefore let’s use the variable x to represent the second zero.
Sum of zeros formula:
5 + x = 7
x = 7 – 5
x = 2
Verification:
– First zero: 5
– Second zero: 2
– Check sum: 5 + 2 = 7
– Check product: 5 * 2 = 10
Mathematical Insight:
Vieta’s formulas offer a beautiful method of determining polynomial zeros
without resorting to complicated solving methods. They show the profound
connection between a polynomial’s coefficients and its roots.
Conclusion:
The other zero of the polynomial is 2.
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The equation x + 2y = 3 can be written in standard form as:
Step 1: Definition of Standard Form Standard form of a linear equation is represented as: Ax + By = C Where: - A, B, and C are constants - A, B, and C are integers - A and B are not both zero - A ≥ 0 (if A = 0, then B must be positive) Step 2: Given Equation Analysis Original equation: x + 2y = 3 StRead more
Step 1: Definition of Standard Form
Standard form of a linear equation is represented as:
Ax + By = C
Where:
– A, B, and C are constants
– A, B, and C are integers
– A and B are not both zero
– A ≥ 0 (if A = 0, then B must be positive)
Step 2: Given Equation Analysis
Original equation: x + 2y = 3
Step 3: Transformation to Standard Form
– The equation is already very close to standard form
– To make it exactly standard form, we need to move all terms to one side
– Rearrange to: x + 2y – 3 = 0
Verification:
– Coefficients of x: 1
– Coefficients of y: 2
– Constant term: -3
– All parts meet standard form requirements
Step 4: Why Other Options Are Incorrect
– “2y + x = 3” ≠ standard form (terms not on one side)
– “3 – x = 2y” ≠ standard form (rearranged incorrectly)
Conclusion:
The correct standard form is x + 2y – 3 = 0
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If a pair of linear equations is inconsistent, then their graphs will be:
Step 1: Definition of Inconsistent Equations - Inconsistent equations are linear equations that have NO SOLUTION - Graphically, this means the lines representing these equations NEVER intersect Step 2: Graphical Interpretation Inconsistent equations always result in PARALLEL LINES - These lines haveRead more
Step 1: Definition of Inconsistent Equations
– Inconsistent equations are linear equations that have NO SOLUTION
– Graphically, this means the lines representing these equations NEVER intersect
Step 2: Graphical Interpretation
Inconsistent equations always result in PARALLEL LINES
– These lines have the same slope but different y-intercepts
– They run alongside each other, maintaining a constant distance
– No point exists where these lines cross
Step 3: Algebraic Characteristics
Example of Inconsistent Equations:
– Equation ₁: 2x + y = 4
– Equation ₂: 2x + y = 6
Observe:
– Same coefficient for x (2)
– Same coefficient for y (1)
– Different constant terms (4 and 6)
– This guarantees the lines will be parallel
Step 4: Geometric Visualization
– Imagine two identical lines shifted vertically
– They have the same “direction” but never touch
– No matter how far you extend them, they remain parallel
Conclusion:
For a pair of linear equations to be inconsistent, their graphs MUST be PARALLEL LINES.
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The pair of equations 2x + 3y = 5** and **4x + 6y = 10 has
Step 1: Equation Analysis Equation ₁: 2x + 3y = 5 Equation ₂: 4x + 6y = 10 Step 2: Proportionality Check - Coefficient of x in Equation ₁: 2 - Coefficient of x in Equation ₂: 4 - Ratio of x coefficients: 4 ÷ 2 = 2 - Coefficient of y in Equation ₁: 3 - Coefficient of y in Equation ₂: 6 - Ratio of y cRead more
Step 1: Equation Analysis
Equation ₁: 2x + 3y = 5
Equation ₂: 4x + 6y = 10
Step 2: Proportionality Check
– Coefficient of x in Equation ₁: 2
– Coefficient of x in Equation ₂: 4
– Ratio of x coefficients: 4 ÷ 2 = 2
– Coefficient of y in Equation ₁: 3
– Coefficient of y in Equation ₂: 6
– Ratio of y coefficients: 6 ÷ 3 = 2
Step 3: Constant Term Verification
– Equation ₁ constant: 5
– Equation ₂ constant: 10
– Ratio of constants: 10 ÷ 5 = 2
Step 4: Algebraic Manipulation
Divide Equation ₂ by 2:
2x + 3y = 5 (Identical to Equation ₁)
Step 5: Interpretation
– The equations represent EXACTLY THE SAME LINE
– This means the system has INFINITELY MANY SOLUTIONS
– Every point on this line satisfies both equations
Mathematical Insight:
When two linear equations represent identical lines,
they have an infinite number of solution points that
perfectly overlap each other.
Conclusion:
The system has INFINITELY MANY SOLUTIONS.
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/