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Expand using the identity (a + b + c) square: (3x – 2y + 4z) square
We expand this expression using the three-term standard identity. Here, our terms are 3x, minus 2y and 4z. Squaring these individual parts results in 9x square, 4y square and 16z square. When calculating the paired products, any term multiplied by minus 2y becomes negative, resulting in minus 12xy aRead more
We expand this expression using the three-term standard identity. Here, our terms are 3x, minus 2y and 4z. Squaring these individual parts results in 9x square, 4y square and 16z square. When calculating the paired products, any term multiplied by minus 2y becomes negative, resulting in minus 12xy and minus 16yz. The final product two times 4z times 3x stays positive at 24zx.
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 lessIs this an identity? (a + b – c) square + (a – b + c) square + (a – b – c) square = 2a square + 2b square + 2c square
To check if this equation is an identity, we must expand all three squared brackets on the left side using the three-term formula. Summing the expanded terms together results in a total of 3a square + 3b square + 3c square minus 2ab minus 2bc minus 2ca. Since this long combined expression is completRead more
To check if this equation is an identity, we must expand all three squared brackets on the left side using the three-term formula. Summing the expanded terms together results in a total of 3a square + 3b square + 3c square minus 2ab minus 2bc minus 2ca. Since this long combined expression is completely different from the given right-hand side of 2a square + 2b square + 2c square, the equation is not an identity.
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 lessFill in the blanks to complete the following identities: s square – 11s + 24 = () ()
To complete this quadratic identity without using tiles, we compare it to the standard form x square + (a + b)x + ab. We look for two integers, a and b, whose sum equals minus 11 and whose product equals 24. The numbers that satisfy both equations are minus 3 and minus 8. Substituting these into theRead more
To complete this quadratic identity without using tiles, we compare it to the standard form x square + (a + b)x + ab. We look for two integers, a and b, whose sum equals minus 11 and whose product equals 24. The numbers that satisfy both equations are minus 3 and minus 8. Substituting these into the factored form results in the final blank answers of (s – 3) and (s – 8).
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 lessFill in the blanks to complete the following identities: (_) (x + 1) = (3x square – 4x – 7)
To find the missing binomial factor, we need to factorize the quadratic expression on the right side. We split the middle coefficient minus 4 into minus 7 and positive 3 because their product matches the product of 3 and minus 7. Rewriting the expression gives 3x square + 3x - 7x - 7. Grouping the tRead more
To find the missing binomial factor, we need to factorize the quadratic expression on the right side. We split the middle coefficient minus 4 into minus 7 and positive 3 because their product matches the product of 3 and minus 7. Rewriting the expression gives 3x square + 3x – 7x – 7. Grouping the terms gives 3x(x + 1) – 7(x + 1). Factoring out the common binomial yields the missing part, (3x – 7).
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 lessFill in the blanks to complete the following identities: 10x square – 11x – 6 = (2x – ) ( + 2)
We solve for the blanks by splitting the middle term of the quadratic expression. We need two numbers that add up to minus 11 and multiply to minus 60. These numbers are minus 15 and positive 4. This transforms the polynomial into 10x square - 15x + 4x - 6. Grouping into pairs gives 5x(2x - 3) + 2(2Read more
We solve for the blanks by splitting the middle term of the quadratic expression. We need two numbers that add up to minus 11 and multiply to minus 60. These numbers are minus 15 and positive 4. This transforms the polynomial into 10x square – 15x + 4x – 6. Grouping into pairs gives 5x(2x – 3) + 2(2x – 3). This gives the complete factored form matching the blanks perfectly.
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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