NCERT Solutions Class 9 Maths Exercise 4.1 — Exploring Algebraic Identities

NCERT Class 9 Mathematics — Ganita Manjari Part I, Exploring Algebraic Identities, Exercise Set 4.1. Both questions solved.

An identity is an equation true for every value of its variables, unlike an ordinary equation which is true only for particular values. The whole chapter rests on that difference, and this exercise uses the first of the identities:

$$(a + b)^2 = a^2 + 2ab + b^2$$

Using it is a matter of deciding what plays the part of $a$ and what plays the part of $b$, then writing down three terms: the square of the first, twice the product, and the square of the second.

Key insight. $a$ and $b$ need not be single letters. They can be $7x$, or $\frac{3}{4}s$, or $\frac{1}{2y}$, or the number $60$ — anything at all. The whole power of an identity is that it holds for every substitution, which is why the same three-term rule expands an algebraic square in Question 1 and squares a three-digit number in Question 2.

Question 1

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$, expand the following:

Solution. In each part, name $a$ and $b$ first, then write the three terms.

(i) $(7x + 4y)^2$

Here $a = 7x$ and $b = 4y$:

$$(7x)^2 + 2(7x)(4y) + (4y)^2 = 49x^2 + 56xy + 16y^2$$

Note that $(7x)^2$ is $49x^2$, not $7x^2$ — the coefficient is squared along with the letter.

(ii) $\left( \dfrac{7}{5}x + \dfrac{3}{2}y \right)^2$

Here $a = \frac{7}{5}x$ and $b = \frac{3}{2}y$:

$$\left(\frac{7}{5}x\right)^2 + 2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right) + \left(\frac{3}{2}y\right)^2$$

$$= \frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2$$

(The middle term: $2 \times \frac{7}{5} \times \frac{3}{2} = \frac{42}{10} = \frac{21}{5}$.)

(iii) $(2.5p + 1.5q)^2$

With $a = 2.5p$ and $b = 1.5q$:

$$(2.5)^2p^2 + 2(2.5)(1.5)pq + (1.5)^2q^2 = 6.25p^2 + 7.5pq + 2.25q^2$$

(iv) $\left( \dfrac{3}{4}s + 8t \right)^2$

$$\left(\frac{3}{4}s\right)^2 + 2\left(\frac{3}{4}s\right)(8t) + (8t)^2 = \frac{9}{16}s^2 + 12st + 64t^2$$

(Middle term: $2 \times \frac{3}{4} \times 8 = 12$.)

(v) $\left( x + \dfrac{1}{2y} \right)^2$

Here $b$ is itself a fraction, $\frac{1}{2y}$:

$$x^2 + 2(x)\left(\frac{1}{2y}\right) + \left(\frac{1}{2y}\right)^2 = x^2 + \frac{x}{y} + \frac{1}{4y^2}$$

(The $2$ cancels against the $2y$ in the middle term; and $\left(\frac{1}{2y}\right)^2 = \frac{1}{4y^2}$, squaring both the $2$ and the $y$.)

(vi) $\left( \dfrac{1}{x} + \dfrac{1}{y} \right)^2$

$$\left(\frac{1}{x}\right)^2 + 2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right) + \left(\frac{1}{y}\right)^2 = \frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}$$

Question 2

Using the same identity, find the values of the following:

Solution. The trick is to split each number into a convenient round part plus a small part, so that $a$ is a multiple of $10$ or $100$ and the squaring is easy.

(i) $(64)^2$

Write $64 = 60 + 4$, so $a = 60$ and $b = 4$:

$$60^2 + 2(60)(4) + 4^2 = 3600 + 480 + 16 = 4096$$

(ii) $(105)^2$

Write $105 = 100 + 5$:

$$100^2 + 2(100)(5) + 5^2 = 10000 + 1000 + 25 = 11025$$

(iii) $(205)^2$

Write $205 = 200 + 5$:

$$200^2 + 2(200)(5) + 5^2 = 40000 + 2000 + 25 = 42025$$

Each of these is three easy multiplications and an addition, all of which can be done in the head — which is exactly what the identity is for.

Common mistakes

• Writing $(a+b)^2 = a^2 + b^2$. This is the single commonest error in algebra and the chapter opens by warning against it. For $(7x + 4y)^2$ it would give $49x^2 + 16y^2$ and lose the $56xy$ entirely. Test it with numbers: $(10+2)^2 = 144$, but $10^2 + 2^2 = 104$.
• Question 1(i), writing $(7x)^2 = 7x^2$. Squaring a product squares both factors: $(7x)^2 = 49x^2$. The same applies to $(4y)^2 = 16y^2$.
• Question 1(ii), forgetting to double. The middle term is $2ab$, so $2 \times \frac{7}{5} \times \frac{3}{2}$, which is $\frac{21}{5}$ and not $\frac{21}{10}$.
• Question 1(v), writing $\left(\frac{1}{2y}\right)^2 = \frac{1}{2y^2}$. Both the $2$ and the $y$ get squared, giving $\frac{1}{4y^2}$.
• Question 1(vi), writing $\frac{1}{x^2 y^2}$ for the middle term. The middle term is $2 \times \frac{1}{x} \times \frac{1}{y} = \frac{2}{xy}$, with $xy$ underneath, not $x^2y^2$.
• Question 2, splitting awkwardly. $64 = 32 + 32$ is valid but gains nothing, since squaring $32$ is no easier than squaring $64$. Choose $a$ to be a round number, so that $a^2$ and $2ab$ are both immediate.
• Question 2(iii), losing a zero in $200^2$. $200^2 = 40{,}000$, not $4{,}000$. Counting the zeros — two in $200$, so four in its square — is a quick guard.

Practise next

• Exercise Set 4.2 — the companion identity $(a – b)^2 = a^2 – 2ab + b^2$, used both to factorise and to square numbers just below a round one.
• The End-of-Chapter Exercises, Questions 1 and 2 — the same two skills with a wider range of identities, including cubes.