NCERT Solutions · Class 9 Mathematics Exploring Algebraic Identities

Choose the identity by looking at the nearest round number. Both identities work for every number, so "easier" simply means: less arithmetic. Go to the nearest number whose square you know at a glance — a multiple of \(\displaystyle 10\), \(\displaystyle 100\) or \(\displaystyle 1000\). If the given number is above it, the gap is added, so use \(\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}\). If it is below it, use \(\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}\). Either way you want the gap \(\displaystyle b\) to be small, because you have to square it.

Number	Best split	Identity that is easier
\(\displaystyle 117^{2}\)	\(\displaystyle 120-3\)	\(\displaystyle (a-b)^{2}\)
\(\displaystyle 78^{2}\)	\(\displaystyle 80-2\)	\(\displaystyle (a-b)^{2}\)
\(\displaystyle 198^{2}\)	\(\displaystyle 200-2\)	\(\displaystyle (a-b)^{2}\)
\(\displaystyle 214^{2}\)	\(\displaystyle 200+14\)	\(\displaystyle (a+b)^{2}\)
\(\displaystyle 1104^{2}\)	\(\displaystyle 1100+4\)	\(\displaystyle (a+b)^{2}\)
\(\displaystyle 1120^{2}\)	\(\displaystyle 1100+20\)	\(\displaystyle (a+b)^{2}\)

(i) \(\displaystyle 117^{2}=(120-3)^{2}=14400-2(120)(3)+9=14400-720+9=13689\)(ii) \(\displaystyle 78^{2}=(80-2)^{2}=6400-2(80)(2)+4=6400-320+4=6084\)(iii) \(\displaystyle 198^{2}=(200-2)^{2}=40000-2(200)(2)+4=40000-800+4=39204\)(iv) \(\displaystyle 214^{2}=(200+14)^{2}=40000+2(200)(14)+196=40000+5600+196=45796\)(v) \(\displaystyle 1104^{2}=(1100+4)^{2}=1210000+2(1100)(4)+16=1210000+8800+16=1218816\)(vi) \(\displaystyle 1120^{2}=(1100+20)^{2}=1210000+2(1100)(20)+400=1210000+44000+400=1254400\)A remark on (iv) and (vi): the choice is not forced. For \(\displaystyle 214\) you could equally take \(\displaystyle 220-6\), giving \(\displaystyle 48400-2640+36=45796\) — the same answer, as it must be. And \(\displaystyle 1120^{2}\) is quickest of all if you notice \(\displaystyle 1120=112\times 10\), so \(\displaystyle 1120^{2}=112^{2}\times 100\). Different routes, one answer; that is the point of an identity.Answers: (i) \(\displaystyle 13689\); (ii) \(\displaystyle 6084\); (iii) \(\displaystyle 39204\); (iv) \(\displaystyle 45796\); (v) \(\displaystyle 1218816\); (vi) \(\displaystyle 1254400\).

Match the shape to the identity. Three-term expressions with two perfect squares point to \(\displaystyle (a\pm b)^{2}=a^{2}\pm 2ab+b^{2}\); six-term expressions with three perfect squares point to \(\displaystyle (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\). In the three-letter case the signs of the cross terms tell you which letters carry a minus sign.(i) \(\displaystyle 16y^{2}-24y+9\). Outer terms: \(\displaystyle (4y)^{2}\) and \(\displaystyle 3^{2}\). The middle term is negative, so try \(\displaystyle (4y-3)^{2}\); check \(\displaystyle 2(4y)(3)=24y\), which matches the \(\displaystyle -24y\). \[16y^{2}-24y+9=(4y-3)^{2} \](ii) \(\displaystyle \frac{9}{4}s^{2}+6st+4t^{2}\). Outer terms: \(\displaystyle \left(\frac{3}{2}s\right)^{2}\) and \(\displaystyle (2t)^{2}\). Check \(\displaystyle 2\left(\frac{3}{2}s\right)(2t)=6st\). It matches. \[\tfrac{9}{4}s^{2}+6st+4t^{2}=\left(\tfrac{3}{2}s+2t\right)^{2}=\tfrac{1}{4}(3s+4t)^{2} \](iii) \(\displaystyle \frac{m^{2}}{9}+\frac{mk}{3}+\frac{k^{2}}{4}+3nk+2mn+9n^{2}\). Six terms, and three of them are squares: \(\displaystyle \frac{m^{2}}{9}=\left(\frac{m}{3}\right)^{2}\), \(\displaystyle \frac{k^{2}}{4}=\left(\frac{k}{2}\right)^{2}\), \(\displaystyle 9n^{2}=(3n)^{2}\). All signs are positive, so try \(\displaystyle a=\frac{m}{3}\), \(\displaystyle b=\frac{k}{2}\), \(\displaystyle c=3n\) and check the three cross terms: \[2ab=2\cdot\tfrac{m}{3}\cdot\tfrac{k}{2}=\tfrac{mk}{3},\qquad 2bc=2\cdot\tfrac{k}{2}\cdot 3n=3nk,\qquad 2ca=2\cdot 3n\cdot\tfrac{m}{3}=2mn. \] All three match the printed terms. \[\tfrac{m^{2}}{9}+\tfrac{mk}{3}+\tfrac{k^{2}}{4}+3nk+2mn+9n^{2}=\left(\tfrac{m}{3}+\tfrac{k}{2}+3n\right)^{2} \](iv) \(\displaystyle \frac{p^{2}}{16}-2+\frac{16}{p^{2}}\). The outer terms are \(\displaystyle \left(\frac{p}{4}\right)^{2}\) and \(\displaystyle \left(\frac{4}{p}\right)^{2}\), and their product is neat: \(\displaystyle 2\cdot\frac{p}{4}\cdot\frac{4}{p}=2\). The middle term is \(\displaystyle -2\), so the sign inside is a minus. \[\tfrac{p^{2}}{16}-2+\tfrac{16}{p^{2}}=\left(\tfrac{p}{4}-\tfrac{4}{p}\right)^{2} \](v) \(\displaystyle 9a^{2}+4b^{2}+c^{2}-12ab+6ac-4bc\). The squares are \(\displaystyle (3a)^{2}\), \(\displaystyle (2b)^{2}\), \(\displaystyle c^{2}\). Two cross terms are negative and one is positive, and both negative ones involve \(\displaystyle b\) — so \(\displaystyle b\) is the term carrying the minus sign. Try \(\displaystyle a\!\to\!3a\), \(\displaystyle b\!\to\!-2b\), \(\displaystyle c\!\to\!c\): \[2(3a)(-2b)=-12ab,\qquad 2(-2b)(c)=-4bc,\qquad 2(c)(3a)=6ac. \] All three match. \[9a^{2}+4b^{2}+c^{2}-12ab+6ac-4bc=(3a-2b+c)^{2} \] (Since squaring kills an overall sign, \(\displaystyle (-3a+2b-c)^{2}\) is the same answer.)Answers: (i) \(\displaystyle (4y-3)^{2}\); (ii) \(\displaystyle \left(\frac{3}{2}s+2t\right)^{2}\); (iii) \(\displaystyle \left(\frac{m}{3}+\frac{k}{2}+3n\right)^{2}\); (iv) \(\displaystyle \left(\frac{p}{4}-\frac{4}{p}\right)^{2}\); (v) \(\displaystyle (3a-2b+c)^{2}\).

Use the identity once for each expression — the only work is deciding what \(\displaystyle a\), \(\displaystyle b\) and \(\displaystyle c\) are.\[(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca \](i) \(\displaystyle (p+3q+7r)^{2}\)Match the expression to the identity: \(\displaystyle a=p\), \(\displaystyle b=3q\), \(\displaystyle c=7r\).Square each term:\[a^{2}=p^{2}, \qquad b^{2}=(3q)^{2}=9q^{2}, \qquad c^{2}=(7r)^{2}=49r^{2} \]Now the three cross terms. Each is twice a product, and the whole coefficient gets squared or multiplied — not just the letter:\[2ab=2(p)(3q)=6pq, \qquad 2bc=2(3q)(7r)=42qr, \qquad 2ca=2(7r)(p)=14rp \]Adding all six:\[(p+3q+7r)^{2}=p^{2}+9q^{2}+49r^{2}+6pq+42qr+14rp \](ii) \(\displaystyle (3x-2y+4z)^{2}\)Here \(\displaystyle a=3x\), \(\displaystyle b=-2y\), \(\displaystyle c=4z\). Take the minus sign into \(\displaystyle b\) rather than trying to remember a different identity — that is the whole trick with a subtraction.Squares (a negative squared is positive, so the \(\displaystyle y^{2}\) term is positive):\[a^{2}=9x^{2}, \qquad b^{2}=(-2y)^{2}=4y^{2}, \qquad c^{2}=16z^{2} \]Cross terms, keeping the sign of \(\displaystyle b\):\[2ab=2(3x)(-2y)=-12xy, \qquad 2bc=2(-2y)(4z)=-16yz, \qquad 2ca=2(4z)(3x)=24zx \]\[(3x-2y+4z)^{2}=9x^{2}+4y^{2}+16z^{2}-12xy-16yz+24zx \]Notice which terms went negative: the two that involve \(\displaystyle y\), because \(\displaystyle y\) is the only one carrying a minus. The \(\displaystyle zx\) term stays positive — two of the three products never touch \(\displaystyle y\).Answer: (i) \(\displaystyle p^{2}+9q^{2}+49r^{2}+6pq+42qr+14rp\) (ii) \(\displaystyle 9x^{2}+4y^{2}+16z^{2}-12xy-16yz+24zx\)

Test a claimed identity by expanding both sides. An identity must be true for every value of every letter. So there are two possible jobs: expand both sides and show they agree, or produce a single set of values where they disagree. Here, expanding shows they do not agree.Expand each square with \(\displaystyle (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\), reading the minus signs into the terms: \[(a+b-c)^{2}=a^{2}+b^{2}+c^{2}+2ab-2bc-2ca \] \[(a-b+c)^{2}=a^{2}+b^{2}+c^{2}-2ab-2bc+2ca \] \[(a-b-c)^{2}=a^{2}+b^{2}+c^{2}-2ab+2bc-2ca \]Now add. The three \(\displaystyle ab\) terms give \(\displaystyle 2ab-2ab-2ab=-2ab\); the three \(\displaystyle bc\) terms give \(\displaystyle -2bc-2bc+2bc=-2bc\); the three \(\displaystyle ca\) terms give \(\displaystyle -2ca+2ca-2ca=-2ca\). So \[(a+b-c)^{2}+(a-b+c)^{2}+(a-b-c)^{2}=3a^{2}+3b^{2}+3c^{2}-2ab-2bc-2ca. \]The right-hand side of the claim is \(\displaystyle 2a^{2}+2b^{2}+2c^{2}\), which is not the same expression.A single counter-example settles it. Take \(\displaystyle a=b=c=1\): \[\text{LHS}=(1+1-1)^{2}+(1-1+1)^{2}+(1-1-1)^{2}=1+1+1=3, \qquad \text{RHS}=2+2+2=6. \] Since \(\displaystyle 3\neq 6\), the statement fails for at least one set of values, so it is not an identity.A caution worth remembering. The equation is not never true — it just is not always true. For instance \(\displaystyle a=0\), \(\displaystyle b=c=1\) gives LHS \(\displaystyle =0+0+4=4\) and RHS \(\displaystyle =0+2+2=4\). Finding values that work is therefore not evidence of an identity; only an expansion valid for all values is. This is exactly why we expand instead of testing a few numbers.Answer: No, it is not an identity. The correct expansion is \(\displaystyle (a+b-c)^{2}+(a-b+c)^{2}+(a-b-c)^{2}=3a^{2}+3b^{2}+3c^{2}-2ab-2bc-2ca\), and \(\displaystyle a=b=c=1\) gives \(\displaystyle 3\) on the left but \(\displaystyle 6\) on the right.Note on the passage that follows the question. The remaining text belongs to the book's discussion, not to the question, and the figure it refers to is not reproduced here. The identity itself needs no figure: starting from \(\displaystyle a^{2}-b^{2}=(a+b)(a-b)\) and adding \(\displaystyle b^{2}\) to both sides gives \[(a+b)(a-b)+b^{2}=a^{2}-b^{2}+b^{2}=a^{2}. \] Śhrīdharāchārya's use of it is to choose \(\displaystyle b\) so that \(\displaystyle a+b\) and \(\displaystyle a-b\) are easy: \(\displaystyle 55^{2}=(55+5)(55-5)+5^{2}=60\times 50+25=3000+25=3025\).