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Using 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/
See lessUsing the identity (a + b) square = a square + 2ab + b square, expand (x + 1/2y) square
We solve this expansion by setting a equal to x and b equal to 1/2y in our identity formula. The square of the first term is x square, and the square of the reciprocal term is 1/4y square. For the middle term, we calculate two times x times 1/2y. During this multiplication, the number two cancels ouRead more
We solve this expansion by setting a equal to x and b equal to 1/2y in our identity formula. The square of the first term is x square, and the square of the reciprocal term is 1/4y square. For the middle term, we calculate two times x times 1/2y. During this multiplication, the number two cancels out completely, leaving x/y as the simplified middle term.
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 (1/x + 1/y) square
This problem requires expanding a expression containing two reciprocal variables. By setting a as 1/x and b as 1/y, we square each part to obtain 1/x square and 1/y square respectively. The middle term of the identity is two times 1/x times 1/y, which merges into the fraction 2/xy. Writing the threeRead more
This problem requires expanding a expression containing two reciprocal variables. By setting a as 1/x and b as 1/y, we square each part to obtain 1/x square and 1/y square respectively. The middle term of the identity is two times 1/x times 1/y, which merges into the fraction 2/xy. Writing the three resulting terms together gives the final complete 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 same identity, find the value of (64) square
To find this value using algebra, we break the number 64 down into 60 + 4. This lets us use the identity with a equal to 60 and b equal to 4. The square of 60 is 3600, and the square of 4 is 16. The middle term is two times 60 times 4, which equals 480. Adding 3600, 480, and 16 together results in 4Read more
To find this value using algebra, we break the number 64 down into 60 + 4. This lets us use the identity with a equal to 60 and b equal to 4. The square of 60 is 3600, and the square of 4 is 16. The middle term is two times 60 times 4, which equals 480. Adding 3600, 480, and 16 together results in 4096.
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 same identity, find the value of (105) square
We evaluate this square by rewriting 105 as the sum of 100 and 5. Applying our identity, we set a to 100 and b to 5. The square of 100 is 10000, and the square of 5 is 25. The cross-product term is calculated as two multiplied by 100 multiplied by 5, which gives 1000. Adding 10000, 1000, and 25 togeRead more
We evaluate this square by rewriting 105 as the sum of 100 and 5. Applying our identity, we set a to 100 and b to 5. The square of 100 is 10000, and the square of 5 is 25. The cross-product term is calculated as two multiplied by 100 multiplied by 5, which gives 1000. Adding 10000, 1000, and 25 together gives 11025.
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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