NCERT Solutions Class 9 Maths Exercise 4.2 — Exploring Algebraic Identities

NCERT Class 9 Mathematics — Ganita Manjari Part I, Exploring Algebraic Identities, Exercise Set 4.2. Both questions solved.

This exercise uses the second identity,

$$(a – b)^2 = a^2 – 2ab + b^2$$

and — more importantly — it runs the first identity backwards. Expanding turns $(a+b)^2$ into three terms; factorising recognises three terms as a square and packs them back up.

The recognition test for $a^2 + 2ab + b^2$:

1. Is the first term a perfect square? If so, $a$ is its square root.
2. Is the last term a perfect square? If so, $b$ is its square root.
3. Is the middle term exactly $2ab$? If yes, the whole thing is $(a+b)^2$.

All three must hold. The third is the one that actually decides it.

Key insight. When the first and last terms are not perfect squares, look for a common factor to take out first. The book’s hint points at this: $50p^2 + 60pq + 18q^2$ becomes $2(25p^2 + 30pq + 9q^2) = 2(5p + 3q)^2$ once the $2$ is removed. Parts (v) and (vi) need the same move with a fraction taken out, which is the only new idea in the exercise.

Question 1

Factor completely:

Solution. Apply the three-part test to each.

(i) $9x^2 + 24xy + 16y^2$

$9x^2 = (3x)^2$ and $16y^2 = (4y)^2$, so try $a = 3x$, $b = 4y$. Check the middle term:

$$2ab = 2(3x)(4y) = 24xy \; \checkmark$$

$$9x^2 + 24xy + 16y^2 = (3x + 4y)^2$$

(ii) $4s^2 + 20st + 25t^2$

$a = 2s$, $b = 5t$, and $2(2s)(5t) = 20st$ ✓

$$4s^2 + 20st + 25t^2 = (2s + 5t)^2$$

(iii) $49x^2 + 28xy + 4y^2$

$a = 7x$, $b = 2y$, and $2(7x)(2y) = 28xy$ ✓

$$49x^2 + 28xy + 4y^2 = (7x + 2y)^2$$

(iv) $64p^2 + \dfrac{32}{3}pq + \dfrac{4}{9}q^2$

The fractions look forbidding but the test is unchanged. $64p^2 = (8p)^2$ and $\frac{4}{9}q^2 = \left(\frac{2}{3}q\right)^2$, so try $a = 8p$, $b = \frac{2}{3}q$:

$$2ab = 2(8p)\left(\frac{2}{3}q\right) = \frac{32}{3}pq \; \checkmark$$

$$64p^2 + \frac{32}{3}pq + \frac{4}{9}q^2 = \left( 8p + \frac{2}{3}q \right)^2$$

This can also be written as $\dfrac{4}{9}(12p + q)^2$, which is the same expression.

(v) $3a^2 + 4ab + \dfrac{4}{3}b^2$

Now $3a^2$ is not a perfect square, so the direct test fails. Following the book’s hint, look for a common factor — here a fractional one. Taking out $\frac{1}{3}$:

$$3a^2 + 4ab + \frac{4}{3}b^2 = \frac{1}{3}\left( 9a^2 + 12ab + 4b^2 \right)$$

(Check each term: $\frac{1}{3} \times 9a^2 = 3a^2$ ✓, $\frac{1}{3} \times 12ab = 4ab$ ✓, $\frac{1}{3} \times 4b^2 = \frac{4}{3}b^2$ ✓.)

The bracket now passes the test with $a’ = 3a$, $b’ = 2b$, since $2(3a)(2b) = 12ab$:

$$3a^2 + 4ab + \frac{4}{3}b^2 = \frac{1}{3}(3a + 2b)^2$$

(vi) $\dfrac{9}{5}s^2 + 6sv + 5v^2$

Same approach. Take out $\frac{1}{5}$:

$$\frac{9}{5}s^2 + 6sv + 5v^2 = \frac{1}{5}\left( 9s^2 + 30sv + 25v^2 \right)$$

(Check: $\frac{1}{5} \times 9s^2 = \frac{9}{5}s^2$ ✓, $\frac{1}{5} \times 30sv = 6sv$ ✓, $\frac{1}{5} \times 25v^2 = 5v^2$ ✓.)

The bracket is a square with $3s$ and $5v$, since $2(3s)(5v) = 30sv$:

$$\frac{9}{5}s^2 + 6sv + 5v^2 = \frac{1}{5}(3s + 5v)^2$$

Question 2

Find the values of the following using the identity $(a – b)^2 = a^2 – 2ab + b^2$:

Solution. Each number is just below a round number, so write it as round minus small. This is why the subtraction identity exists: it turns awkward numbers into easy ones.

(i) $(79)^2$

$79 = 80 – 1$, so $a = 80$, $b = 1$:

$$80^2 – 2(80)(1) + 1^2 = 6400 – 160 + 1 = 6241$$

(ii) $(193)^2$

$193 = 200 – 7$:

$$200^2 – 2(200)(7) + 7^2 = 40000 – 2800 + 49 = 37249$$

(iii) $(299)^2$

$299 = 300 – 1$:

$$300^2 – 2(300)(1) + 1^2 = 90000 – 600 + 1 = 89401$$

Compare the work with the addition identity: $79 = 70 + 9$ would give $4900 + 1260 + 81$, which is correct but considerably more effort. Choosing the nearest round number is what makes the calculation short, and that is what decides whether to use $(a+b)^2$ or $(a-b)^2$.

Common mistakes

• Checking only the first and last terms. $9x^2 + 20xy + 16y^2$ has perfect squares at both ends but is not $(3x+4y)^2$, because $2(3x)(4y) = 24xy$, not $20xy$. The middle term is the test that matters.
• Question 1(iv), writing $\left(\frac{2}{3}q\right)^2 = \frac{2}{3}q^2$. Squaring a fraction squares top and bottom: $\frac{4}{9}q^2$.
• Question 1(v), declaring it does not factor. $3a^2$ is not a perfect square, so the direct test fails — but taking out $\frac{1}{3}$ first makes everything square. The book’s hint about Example 7 is pointing exactly here.
• Question 1(v) and (vi), taking out the wrong factor. Pulling out $3$ from $3a^2 + 4ab + \frac{4}{3}b^2$ gives $3\left(a^2 + \frac{4}{3}ab + \frac{4}{9}b^2\right)$, which is also correct but leaves fractions inside. Taking out $\frac{1}{3}$ clears them. Always multiply back to check whichever you choose.
• Question 2, using the wrong identity. For $299$, the nearest round number is $300$ and the difference is $1$. Treating it as $(290 + 9)^2$ works but replaces a subtraction of $600$ with a multiplication by $9$.
• Question 2(ii), forgetting the sign of the middle term. $(a-b)^2$ subtracts $2ab$. Adding it gives $40000 + 2800 + 49 = 42849$, which is $207^2$ — the square of the wrong number.
• Question 2, forgetting to square $b$. The last term is $b^2$, not $b$: for $193^2$ it is $49$, not $7$.

Practise next

• Exercise Set 4.3 — a third identity, $(a+b+c)^2$, together with questions asking you to choose which of the three identities makes a calculation easiest.
• The End-of-Chapter Exercises, Question 3 — a long list of factorisations of exactly this kind, including several needing a common factor taken out first.