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Factor completely 3a square + 4ab + 4/3 b square
For this problem, we take out a fractional common factor to make the terms perfect integers. Factoring out 1/3 transforms the expression into 1/3 times the quantity 9a square + 12ab + 4b square. Inside the brackets, 9a square is the square of 3a, and 4b square is the square of 2b. The middle term isRead more
For this problem, we take out a fractional common factor to make the terms perfect integers. Factoring out 1/3 transforms the expression into 1/3 times the quantity 9a square + 12ab + 4b square. Inside the brackets, 9a square is the square of 3a, and 4b square is the square of 2b. The middle term is two times 3a times 2b. This simplifies perfectly to 1/3 times (3a + 2b) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessFactor completely 9/5 s square + 6sv + 5v square
We simplify this factorization by removing 1/5 as a common factor first. This leaves us with 1/5 multiplied by the expression 9s square + 30sv + 25v square. Now we can easily see that 9s square is the square of 3s, and 25v square is the square of 5v. The middle term matches two times 3s times 5v. ThRead more
We simplify this factorization by removing 1/5 as a common factor first. This leaves us with 1/5 multiplied by the expression 9s square + 30sv + 25v square. Now we can easily see that 9s square is the square of 3s, and 25v square is the square of 5v. The middle term matches two times 3s times 5v. The fully factored final form is 1/5 times (3s + 5v) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessUsing the identity (a – b) square = a square – 2ab + b square, find the value of (79) square
To calculate this square easily without direct vertical multiplication, we express 79 as 80 - 1. This lets us use the subtraction square identity where a is 80 and b is 1. The square of 80 is 6400, and the square of 1 is 1. The middle term to subtract is two times 80 times 1, which is 160. ComputingRead more
To calculate this square easily without direct vertical multiplication, we express 79 as 80 – 1. This lets us use the subtraction square identity where a is 80 and b is 1. The square of 80 is 6400, and the square of 1 is 1. The middle term to subtract is two times 80 times 1, which is 160. Computing 6400 minus 160 plus 1 gives 6241.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessUsing the identity (a – b) square = a square – 2ab + b square, find the value of (193) square
We find the value by breaking 193 down into the expression 200 - 7. Using our identity, we assign a to be 200 and b to be 7. The square of 200 results in 40000, and the square of 7 is 49. The cross-product term to subtract is calculated as two multiplied by 200 multiplied by 7, which equals 2800. SuRead more
We find the value by breaking 193 down into the expression 200 – 7. Using our identity, we assign a to be 200 and b to be 7. The square of 200 results in 40000, and the square of 7 is 49. The cross-product term to subtract is calculated as two multiplied by 200 multiplied by 7, which equals 2800. Subtracting 2800 from 40000 and adding 49 yields 37249.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessFind the square of 117 using a suitable identity
To calculate the square of 117 efficiently, we split the number into three manageable parts: 100 + 10 + 7. This allows us to use the algebraic identity for a three-term addition square, where a equals 100, b equals 10, and c equals 7. We find the squares of these numbers, which are 10000, 100, and 4Read more
To calculate the square of 117 efficiently, we split the number into three manageable parts: 100 + 10 + 7. This allows us to use the algebraic identity for a three-term addition square, where a equals 100, b equals 10, and c equals 7. We find the squares of these numbers, which are 10000, 100, and 49. Adding these to the cross-product terms 2000, 1400, and 140 results in 13689.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
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