{"id":31499,"date":"2026-05-12T17:45:34","date_gmt":"2026-05-12T12:15:34","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31499"},"modified":"2026-05-12T17:46:45","modified_gmt":"2026-05-12T12:16:45","slug":"exploring-algebraic-identities-ncert-solutions-class-9-ganita-manjari","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/exploring-algebraic-identities-ncert-solutions-class-9-ganita-manjari\/","title":{"rendered":"Exploring Algebraic Identities – NCERT Solutions Class 9 Ganita Manjari"},"content":{"rendered":"\n

Exploring Algebraic Identities – NCERT Solutions Class 9 Maths Ganita Manjari includes all the questions with solutions given in NCERT Class 9 Ganita Manjari<\/a> textbook.<\/p>\n\n\n\n

NCERT Solutions Class 9<\/h2>\n\n\nEnglish Kaveri<\/a>\n\n\nHindi Ganga<\/a>\n\n\nSanskrit Sharada<\/a>\n\n\nMaths Ganita Manjari<\/a>\n\n\nScience Exploration<\/a>\n\n\nSocial Understanding Society<\/a>\n\n\n\n

Exploring Algebraic Identities \u2013 NCERT Solutions<\/h2>\n\n\n\n
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Q.1:<\/p>\n\n\n\n

What can you say about a and b if (a + b)2<\/sup> < a2<\/sup> + b2<\/sup>?<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

{tex} (a+b)^2<a^2+b^2 {\/tex}
Expand LHS:<\/p>\n\n\n\n

{tex} (a+b)^2=a^2+2 a b+b^2 {\/tex}
So inequality becomes:<\/p>\n\n\n\n

{tex} a^2+2 a b+b^2<a^2+b^2 {\/tex}
Subtract {tex}a^2+b^2{\/tex} from both sides:<\/p>\n\n\n\n

{tex} 2 a b <0 {\/tex}
{tex} a b <0 {\/tex}
Conclusion:
{tex}a{\/tex} and {tex}b{\/tex} have opposite signs
(one is positive and the other is negative)<\/p>\n\n\n\n

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Q.2:<\/p>\n\n\n\n

What can you say about a and b if (a + b)2<\/sup> > a2<\/sup> + b2<\/sup>?<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

{tex} (a+b)^2=a^2+2 a b+b^2 {\/tex}<\/p>\n\n\n\n

a = 8
b = 4
{tex} (a+b)^2=a^2+2 a b+b^2 {\/tex}
{tex} 12^2=64+32+32+16=144 {\/tex}

Comparing with {tex}a^2+b^2{\/tex}:<\/p>\n\n\n\n

{tex} (a+b)^2=a^2+b^2+2 a b {\/tex}
So,<\/p>\n\n\n\n

{tex} (a+b)^2>a^2+b^2 \\Rightarrow 2 a b>0 \\Rightarrow a b>0 {\/tex}
This means:
{tex} a{\/tex} and {tex}b{\/tex} have the same sign (both positive or both negative)<\/p>\n\n\n\n

Therefore, when {tex}a{\/tex} and {tex}b{\/tex} are of the same sign, {tex}(a+b)^2>a^2+b^2{\/tex}.<\/p>\n\n\n\n

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Q.3:<\/p>\n\n\n\n

When will (a + b)2<\/sup> be equal to a2<\/sup> + b2<\/sup>?<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

For {tex}(a+b)^2{\/tex} to be equal to {tex}a^2+b^2{\/tex}:<\/p>\n\n\n\n

{tex} a^2+2 a b+b^2=a^2+b^2 {\/tex}
{tex} \\Rightarrow 2 a b=0 \\Rightarrow a b=0 {\/tex}
This happens when:
{tex}a=0{\/tex} or {tex}b=0{\/tex}<\/p>\n\n\n\n

Therefore, {tex}(a+b)^2=a^2+b^2{\/tex} when one of the numbers is zero.<\/p>\n\n\n\n

\n\n\n\n

Q.4:<\/p>\n\n\n\n

Did you observe that {tex}(a+b)^2{\/tex} and {tex}a^2+b^2{\/tex} are both positive? What term will decide which is larger? Use the expansion of {tex}(a+b)^2{\/tex} to decide.<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Both {tex}(a+b)^2{\/tex} and {tex}a^2+b^2{\/tex} are always positive (since they are sums of squares).
Now compare them:<\/p>\n\n\n\n

{tex} (a+b)^2=a^2+b^2+2 a b {\/tex}
So, the term that decides which is larger is {tex}2 a b{\/tex}.<\/p>\n\n\n\n