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Factor using suitable identities: 9/4 s square + 6st + 4t square
To factor this expression, we examine the perfect square parts at the ends. The first term 9/4 s square is equal to the quantity 3/2 s squared. The third term 4t square is equal to the quantity 2t squared. The middle term 6st is exactly equal to two times 3/2 s times 2t. This matches the standard adRead more
To factor this expression, we examine the perfect square parts at the ends. The first term 9/4 s square is equal to the quantity 3/2 s squared. The third term 4t square is equal to the quantity 2t squared. The middle term 6st is exactly equal to two times 3/2 s times 2t. This matches the standard addition identity, resulting in the factor (3/2 s + 2t) 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, expand (7x + 4y) square
To expand this expression, we use the algebraic identity for the square of a sum. We assign the value 7x to a and 4y to b. Squaring 7x gives 49x square, and squaring 4y gives 16y square. The middle term is found by multiplying two by 7x and 4y, which results in 56xy. Combining these three parts giveRead more
To expand this expression, we use the algebraic identity for the square of a sum. We assign the value 7x to a and 4y to b. Squaring 7x gives 49x square, and squaring 4y gives 16y square. The middle term is found by multiplying two by 7x and 4y, which results in 56xy. Combining these three parts gives the final expanded polynomial expression.
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, expand (7/5 x + 3/2 y) square
We expand this fractional binomial using the standard addition square formula. Here, our first term a is 7/5 x and our second term b is 3/2 y. The square of 7/5 x is 49/25 x square, and the square of 3/2 y is 9/4 y square. The middle term is calculated as two times 7/5 x times 3/2 y, which simplifieRead more
We expand this fractional binomial using the standard addition square formula. Here, our first term a is 7/5 x and our second term b is 3/2 y. The square of 7/5 x is 49/25 x square, and the square of 3/2 y is 9/4 y square. The middle term is calculated as two times 7/5 x times 3/2 y, which simplifies cleanly to 21/5 xy.
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, expand (2.5p + 1.5q) square
This expression is expanded by substituting decimal coefficients into the identity formula. We set a equal to 2.5p and b equal to 1.5q. Squaring 2.5p results in 6.25p square, while squaring 1.5q results in 2.25q square. Multiplying two by 2.5p and 1.5q provides the middle term, which is 7.5pq. CombiRead more
This expression is expanded by substituting decimal coefficients into the identity formula. We set a equal to 2.5p and b equal to 1.5q. Squaring 2.5p results in 6.25p square, while squaring 1.5q results in 2.25q square. Multiplying two by 2.5p and 1.5q provides the middle term, which is 7.5pq. Combining these decimal values completes the expansion process.
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, expand (3/4 s + 8t) square
To expand this binomial, we apply the standard square formula where a represents 3/4 s and b represents 8t. Squaring the first term results in 9/16 s square, and squaring the second term yields 64t square. The middle cross-product term is calculated by multiplying two times 3/4 s times 8t, which simRead more
To expand this binomial, we apply the standard square formula where a represents 3/4 s and b represents 8t. Squaring the first term results in 9/16 s square, and squaring the second term yields 64t square. The middle cross-product term is calculated by multiplying two times 3/4 s times 8t, which simplifies directly to the integer 12st. Putting all parts together gives the answer.
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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