If only (a^2+b^2+c^2) is written for ( (a+b+c)^2 ), which terms are missing?
The square of three terms also includes twice the pair products. Exam tip: always write (2ab+2bc+2ca).
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SubjectsMathematics
बीजीय सर्वसमिकाएँ
In this Class 9 Mathematics topic from “Exploring Algebraic Identities,” students learn how standard algebraic relationships remain true for all permitted values of the variables. They study and apply identities such as (a + b)², (a − b)², and a² − b² to expand expressions, simplify calculations, and factorise algebraic forms. The topic also develops skill in recognising suitable patterns, substituting values to verify results, and using identities to solve expressions efficiently and accurately.
TOPIC PRACTICE
Up to 20 questions from this page. Select your focus, then start.
The square of three terms also includes twice the pair products. Exam tip: always write (2ab+2bc+2ca).
View question details\(x^2+12x+36=(x+6)^2\) because the middle term is \(2\times x\times6=12x\). In option B, the constant term should be \(36\), not \(18\). Exam tip: check whether \(b^2=4ac\).
View question detailsHere, \(51=50+1\) and \(49=50-1\). Using the identity \((a+b)^2+(a-b)^2=2a^2+2b^2\), with \(a=50\) and \(b=1\), we get \(51^2+49^2=2(50)^2+2(1)^2=5000+2=5002\). The value 5000 includes only \(2\times 50^2\); \(2\times 1^2\) must also be added. Exam tip: apply this identity directly when two numbers are equally spaced from a middle number.
View question detailsUse the identity \(a^2-b^2=(a+b)(a-b)\). Here, \(51^2-49^2=(51+49)(51-49)=100\times2=200\). Option 100 is only the sum of the two numbers, while 2 is their difference; their product gives the required value. Exam tip: for the difference of squares of nearby numbers, apply this identity directly.
View question detailsUsing \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\), we get \(a^2+b^2+c^2=12^2-2\times35=144-70=74\). Thus, the claim is correct. \(109\) results from forgetting the factor \(2\). Exam tip: always check the coefficient of mixed terms.
View question detailsFirst, use the identity \((x-1)(x+1)=x^2-1\). The expression then becomes \((x^2-1)(x^2+1)\), which again has the form \((a-b)(a+b)=a^2-b^2\). Therefore, it simplifies to \(x^4-1\). \(x^4+1\) is not the result of this difference-of-squares identity. Exam tip: identify conjugate binomial pairs and apply identities step by step.
View question detailsThis is a difference of squares: \(a^2-b^2=(a+b)(a-b)\). Here, \(a=2x+3\) and \(b=2x+1\). Thus, \(a+b=4x+4\) and \(a-b=2\), so the expression becomes \((4x+4)\times2=8x+8\). The expression \(4x+4\) is only \(a+b\), not the final simplified form. Exam tip: first identify the two bases of the squares, then apply the difference-of-squares identity.
View question detailsUse the identity \(a^2-b^2=(a+b)(a-b)\). Here, \(a=3x+5\) and \(b=3x-1\). Thus, \(a+b=6x+4\) and \(a-b=6\). Therefore, \((6x+4)\times6=36x+24\). Option B results from an incorrect multiplication of the \(6x\) term by 6. Exam tip: identify the two squared expressions first and apply the identity directly instead of expanding both squares.
View question detailsHere \(9p^2=(3p)^2\) and \(16q^2=(4q)^2\). The middle term is \(-24pq=-2(3p)(4q)\), so the expression is \((3p-4q)^2\). Exam tip: check the sign of the middle term carefully.
View question detailsGiven \(a+b+c=10\), we get \((a+b+c)^2=10^2=100\). Option 10 is only the value of the sum, not its square. Exam tip: when a power is outside brackets, substitute the bracket's value first and then apply the power.
View question detailsHere \(4a^2=(2a)^2\) and \(9b^2=(3b)^2\). The required middle term is \(2\times2a\times3b=12ab\), so the expression is \((2a+3b)^2\). Exam tip: always check both the sign and coefficient of the middle term.
View question detailsSquaring a sum gives the square of the first term, twice their product, and the square of the second term. Hence \((a+b)^2=a^2+2ab+b^2\). Exam tip: the middle term is \(+2ab\) for a sum, not \(-2ab\).
View question detailsFirst identify (x^2+6x+9=(x+3)^2). Then apply the difference of squares identity to ( (x+3)^2-y^2 ).
View question detailsIn \(x^2+14x+49\), the last term is \(49=7^2\) and the middle term is \(14x=2\cdot x\cdot7\). Hence it is \((x+7)^2\). In exams, always check whether the middle term equals \(2ab\).
View question detailsThis is the form \( (u+v)(u-v)=u^2-v^2 \). With \(u=3a\) and \(v=4b\), it gives \(9a^2-16b^2\).
View question detailsUse the identity \((a+b)^2=a^2+2ab+b^2\). Here, \(x^2+12x+36=x^2+2\cdot x\cdot6+6^2\), so it equals \((x+6)^2\). Note that \((x-6)^2=x^2-12x+36\), so it is not correct. Exam tip: use the sign of the middle term to decide whether the binomial has \(+\) or \(-\).
View question detailsHere, \(96=100-4\) and \(104=100+4\). Therefore, \((100-4)(100+4)=100^2-4^2=10000-16=9984\). The option \(10016\) results from adding \(4^2\) instead of subtracting it. Exam tip: when two numbers are equally spaced around a middle number, use \((a-b)(a+b)=a^2-b^2\).
View question detailsUsing \((a-b)^2=a^2-2ab+b^2\), put \(a=p\) and \(b=q\) to get \(p^2-2pq+q^2\). Option C is the difference of squares. Exam tip: the middle term must be \(-2pq\).
View question details\((x+y)^2=(x+y)(x+y)\). On multiplying, the middle terms give \(xy+xy=2xy\), so the expression is \(x^2+2xy+y^2\). Exam tip: check the sign of the middle term carefully.
View question detailsThis is a difference of squares: \(36-x^2=6^2-x^2\). Using \(a^2-b^2=(a+b)(a-b)\), we get \((6+x)(6-x)\). Option A expands to \(x^2-36\), which is the negative of the given expression. Exam tip: first identify the square roots of both terms, then write one sum factor and one difference factor.
View question detailsQUIZ COMPLETE