NCERT Solutions Class 9 Maths Exercise 4.3 — Exploring Algebraic Identities

NCERT Class 9 Mathematics — Ganita Manjari Part I, Exploring Algebraic Identities, Exercise Set 4.3. All 4 questions solved.

The chapter now has three identities in hand:

$$(a+b)^2 = a^2 + 2ab + b^2$$ $$(a-b)^2 = a^2 – 2ab + b^2$$ $$(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$

Question 1 asks not just for answers but for a choice — which identity makes each calculation easiest. Question 4 asks whether a proposed identity is genuine, which is a different kind of question again: one where the answer may be no.

Key insight. Deciding which identity to use is really about finding the nearest convenient number. $117$ is $100 + 17$ but also $100 + 10 + 7$; the second splits into three easy pieces and uses the three-term identity. $198$ is $200 – 2$, close to a round number, so the subtraction identity wins. The identity that makes the arithmetic easy is the one whose pieces are round.

Question 1

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

Solution.

(i) $117^2$

Split as $100 + 10 + 7$ and use the three-term identity:

$$117^2 = 100^2 + 10^2 + 7^2 + 2(100)(10) + 2(10)(7) + 2(7)(100)$$ $$= 10000 + 100 + 49 + 2000 + 140 + 1400 = 13689$$

(ii) $78^2$

$78 = 80 – 2$, so the subtraction identity:

$$78^2 = 80^2 – 2(80)(2) + 2^2 = 6400 – 320 + 4 = 6084$$

(iii) $198^2$

$198 = 200 – 2$ — subtraction identity again:

$$198^2 = 200^2 – 2(200)(2) + 2^2 = 40000 – 800 + 4 = 39204$$

(iv) $214^2$

$214 = 200 + 14$ — addition identity:

$$214^2 = 200^2 + 2(200)(14) + 14^2 = 40000 + 5600 + 196 = 45796$$

(v) $1104^2$

$1104 = 1100 + 4$ — addition identity:

$$1104^2 = 1100^2 + 2(1100)(4) + 4^2 = 1210000 + 8800 + 16 = 1218816$$

(vi) $1120^2$

$1120 = 1100 + 20$ — addition identity:

$$1120^2 = 1100^2 + 2(1100)(20) + 20^2 = 1210000 + 44000 + 400 = 1254400$$

Which identity when. The rule that emerges: if the number is just below a round number, use $(a-b)^2$; just above, use $(a+b)^2$; and if it is far from any round number but close to a sum of round pieces, use $(a+b+c)^2$. For $117$ the three-term split avoids having to square $17$.

Question 2

Factor using suitable identities:

Solution.

(i) $16y^2 – 24y + 9$

The middle sign is negative, so try $(a-b)^2$ with $a = 4y$ and $b = 3$. Check: $2(4y)(3) = 24y$ ✓

$$16y^2 – 24y + 9 = (4y – 3)^2$$

(ii) $\dfrac{9}{4}s^2 + 6st + 4t^2$

$\frac{9}{4}s^2 = \left(\frac{3}{2}s\right)^2$ and $4t^2 = (2t)^2$; the middle term $2\left(\frac{3}{2}s\right)(2t) = 6st$ ✓

$$\frac{9}{4}s^2 + 6st + 4t^2 = \left( \frac{3}{2}s + 2t \right)^2 = \frac{1}{4}(3s + 4t)^2$$

(iii) $\dfrac{m^2}{9} + \dfrac{mk}{3} + \dfrac{k^2}{4} + 3nk + 2mn + 9n^2$

Six terms means the three-term identity. Three of the terms are squares:

$$\frac{m^2}{9} = \left(\frac{m}{3}\right)^2, \qquad \frac{k^2}{4} = \left(\frac{k}{2}\right)^2, \qquad 9n^2 = (3n)^2$$

So try $a = \frac{m}{3}$, $b = \frac{k}{2}$, $c = 3n$ and check the three cross terms:

$$2ab = 2\left(\frac{m}{3}\right)\left(\frac{k}{2}\right) = \frac{mk}{3} \; \checkmark$$ $$2bc = 2\left(\frac{k}{2}\right)(3n) = 3kn \; \checkmark$$ $$2ca = 2(3n)\left(\frac{m}{3}\right) = 2mn \; \checkmark$$

All three match, so

$$= \left( \frac{m}{3} + \frac{k}{2} + 3n \right)^2$$

(iv) $\dfrac{p^2}{16} – 2 + \dfrac{16}{p^2}$

$\frac{p^2}{16} = \left(\frac{p}{4}\right)^2$ and $\frac{16}{p^2} = \left(\frac{4}{p}\right)^2$. The middle term should be $-2ab$:

$$2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right) = 2 \; \checkmark$$

(the $p$s and the $4$s cancel completely). So

$$\frac{p^2}{16} – 2 + \frac{16}{p^2} = \left( \frac{p}{4} – \frac{4}{p} \right)^2$$

The bare $-2$ in the middle, with no letters at all, is what makes this one look strange — and it is exactly the signal that the two outer terms are reciprocals of each other.

(v) $9a^2 + 4b^2 + c^2 – 12ab + 6ac – 4bc$

Six terms again, with squares $(3a)^2$, $(2b)^2$, $c^2$. The signs of the cross terms tell you which letters carry a minus. Try $a’ = 3a$, $b’ = -2b$, $c’ = c$:

$$2a’b’ = 2(3a)(-2b) = -12ab \; \checkmark$$ $$2b’c’ = 2(-2b)(c) = -4bc \; \checkmark$$ $$2c’a’ = 2(c)(3a) = 6ac \; \checkmark$$

All three match, so

$$9a^2 + 4b^2 + c^2 – 12ab + 6ac – 4bc = (3a – 2b + c)^2$$

Question 3

Expand the following using the identity $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$:

Solution. Six terms every time: three squares and three doubled products.

(i) $(p + 3q + 7r)^2$

With $a = p$, $b = 3q$, $c = 7r$:

$$p^2 + (3q)^2 + (7r)^2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)$$ $$= p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr$$

(ii) $(3x – 2y + 4z)^2$

Treat the minus as part of the letter: $a = 3x$, $b = -2y$, $c = 4z$. The squares are all positive, but the cross terms carry signs.

$$(3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x)$$ $$= 9x^2 + 4y^2 + 16z^2 – 12xy – 16yz + 24xz$$

Note $(-2y)^2 = +4y^2$: squaring destroys the sign, so only the cross terms can be negative.

Question 4

Is this an identity?

$$(a + b – c)^2 + (a – b + c)^2 + (a – b – c)^2 = 2a^2 + 2b^2 + 2c^2$$

Solution. An identity must hold for every value of the variables, so there are two ways to settle it: expand both sides and compare, or find a single set of values where it fails. Do both.

Expand the left-hand side. Using the three-term identity on each bracket:

$$(a + b – c)^2 = a^2 + b^2 + c^2 + 2ab – 2bc – 2ca$$ $$(a – b + c)^2 = a^2 + b^2 + c^2 – 2ab – 2bc + 2ca$$ $$(a – b – c)^2 = a^2 + b^2 + c^2 – 2ab + 2bc – 2ca$$

Adding, the squares give $3a^2 + 3b^2 + 3c^2$, and the cross terms give

$$ab: \; 2 – 2 – 2 = -2 \qquad bc: \; -2 – 2 + 2 = -2 \qquad ca: \; -2 + 2 – 2 = -2$$

So the left-hand side is

$$3a^2 + 3b^2 + 3c^2 – 2ab – 2bc – 2ca$$

which is not $2a^2 + 2b^2 + 2c^2$: the squares have coefficient $3$ rather than $2$, and there are three cross terms on the left and none on the right.

A counterexample. Take $a = b = c = 1$:

$$\text{LHS} = (1+1-1)^2 + (1-1+1)^2 + (1-1-1)^2 = 1^2 + 1^2 + (-1)^2 = 3$$ $$\text{RHS} = 2 + 2 + 2 = 6$$

$3 \neq 6$, so the statement fails, and one failure is enough to disqualify it.

What would make it true. Adding a fourth bracket, $(a + b + c)^2$, gives

$$(a+b+c)^2 + (a+b-c)^2 + (a-b+c)^2 + (a-b-c)^2 = 4a^2 + 4b^2 + 4c^2$$

because the four sets of cross terms then cancel exactly. That is an identity, and it shows what went wrong: the three brackets given are not a symmetric set, so their cross terms do not cancel.

Common mistakes

• Question 1, using the same identity every time. The question explicitly asks which makes each calculation easier. Splitting $198$ as $190 + 8$ gives the right answer with far more work than $200 – 2$.
• Question 1(i), splitting $117$ as $100 + 17$. Correct, but then $17^2$ and $2 \times 100 \times 17$ must be computed. Three round pieces avoid both.
• Question 2(iv), being thrown by the bare $-2$. A middle term with no letters is the clue that the outer terms are reciprocals, since $\frac{p}{4} \times \frac{4}{p} = 1$ and twice that is $2$.
• Question 2(v), assigning signs to the squares. $(-2b)^2$ is $+4b^2$. Only the cross terms can be negative, so the signs of the cross terms are what reveal which letters carry the minus.
• Question 3(ii), writing $(-2y)^2 = -4y^2$. The square of a negative is positive. If a squared term comes out negative, a sign has been dropped.
• Question 4, answering “yes” after checking one case. Try $a = 1$, $b = c = 0$: the left gives $1 + 1 + 1 = 3$ and the right gives $2$ — but $a = 1$, $b = 1$, $c = 0$ gives $4$ on the left and $4$ on the right. One case agreeing proves nothing; an identity must hold for all values, so it has to be expanded.
• Question 4, concluding the question is misprinted. It is not. “Is this an identity?” is a genuine question, and no is a complete and correct answer when supported by an expansion or a counterexample.

Practise next

• Exercise Set 4.4 — factorising quadratics that are not perfect squares, by splitting the middle term.
• The End-of-Chapter Exercises, Question 3(x) — a six-term expression of exactly the shape of Question 2(iii) and (v) here, and worth attempting straight after them.