Exercise 4.1 Practice
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Overview
This page provides comprehensive Ch 4: Algebraic Identities – Exercise 4.1 Practice. Practice expanding expressions using the identity $(a + b)^2 = a^2 + 2ab + b^2$, and apply it to evaluate squares of large numbers mentally. Free step-by-step solutions and explanations.
Algebraic Identity: $(a + b)^2 = a^2 + 2ab + b^2$
Q1: Expand Using (a + b)²
Using the identity $(a + b)^2 = a^2 + 2ab + b^2$, expand the following:
(i) $(7x + 4y)^2$ (ii) $\left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2$ (iii) $(2.5p + 1.5q)^2$
(iv) $\left(\dfrac{3}{4}s + 8t\right)^2$ (v) $\left(x + \dfrac{1}{2y}\right)^2$ (vi) $\left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2$
(i) $(7x + 4y)^2$ (ii) $\left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2$ (iii) $(2.5p + 1.5q)^2$
(iv) $\left(\dfrac{3}{4}s + 8t\right)^2$ (v) $\left(x + \dfrac{1}{2y}\right)^2$ (vi) $\left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2$
Identity: $(a + b)^2 = a^2 + 2ab + b^2$
(i) $(7x + 4y)^2$
Here $a = 7x$, $b = 4y$. $$= (7x)^2 + 2(7x)(4y) + (4y)^2$$ $$= 49x^2 + 56xy + 16y^2$$
Here $a = 7x$, $b = 4y$. $$= (7x)^2 + 2(7x)(4y) + (4y)^2$$ $$= 49x^2 + 56xy + 16y^2$$
(ii) $\left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2$
Here $a = \dfrac{7}{5}x$, $b = \dfrac{3}{2}y$. $$= \left(\frac{7}{5}x\right)^2 + 2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right) + \left(\frac{3}{2}y\right)^2$$ $$= \frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2$$
Here $a = \dfrac{7}{5}x$, $b = \dfrac{3}{2}y$. $$= \left(\frac{7}{5}x\right)^2 + 2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right) + \left(\frac{3}{2}y\right)^2$$ $$= \frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2$$
(iii) $(2.5p + 1.5q)^2$
Here $a = 2.5p$, $b = 1.5q$. $$= (2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2$$ $$= 6.25p^2 + 7.5pq + 2.25q^2$$
Here $a = 2.5p$, $b = 1.5q$. $$= (2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2$$ $$= 6.25p^2 + 7.5pq + 2.25q^2$$
(iv) $\left(\dfrac{3}{4}s + 8t\right)^2$
Here $a = \dfrac{3}{4}s$, $b = 8t$. $$= \left(\frac{3}{4}s\right)^2 + 2\left(\frac{3}{4}s\right)(8t) + (8t)^2$$ $$= \frac{9}{16}s^2 + 12st + 64t^2$$
Here $a = \dfrac{3}{4}s$, $b = 8t$. $$= \left(\frac{3}{4}s\right)^2 + 2\left(\frac{3}{4}s\right)(8t) + (8t)^2$$ $$= \frac{9}{16}s^2 + 12st + 64t^2$$
(v) $\left(x + \dfrac{1}{2y}\right)^2$
Here $a = x$, $b = \dfrac{1}{2y}$. $$= x^2 + 2(x)\left(\frac{1}{2y}\right) + \left(\frac{1}{2y}\right)^2$$ $$= x^2 + \frac{x}{y} + \frac{1}{4y^2}$$
Here $a = x$, $b = \dfrac{1}{2y}$. $$= x^2 + 2(x)\left(\frac{1}{2y}\right) + \left(\frac{1}{2y}\right)^2$$ $$= x^2 + \frac{x}{y} + \frac{1}{4y^2}$$
(vi) $\left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2$
Here $a = \dfrac{1}{x}$, $b = \dfrac{1}{y}$. $$= \left(\frac{1}{x}\right)^2 + 2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right) + \left(\frac{1}{y}\right)^2$$ $$= \frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}$$
Here $a = \dfrac{1}{x}$, $b = \dfrac{1}{y}$. $$= \left(\frac{1}{x}\right)^2 + 2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right) + \left(\frac{1}{y}\right)^2$$ $$= \frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}$$
(i) $49x^2 + 56xy + 16y^2$
(ii) $\dfrac{49}{25}x^2 + \dfrac{21}{5}xy + \dfrac{9}{4}y^2$
(iii) $6.25p^2 + 7.5pq + 2.25q^2$
(iv) $\dfrac{9}{16}s^2 + 12st + 64t^2$ (v) $x^2 + \dfrac{x}{y} + \dfrac{1}{4y^2}$ (vi) $\dfrac{1}{x^2} + \dfrac{2}{xy} + \dfrac{1}{y^2}$
(iv) $\dfrac{9}{16}s^2 + 12st + 64t^2$ (v) $x^2 + \dfrac{x}{y} + \dfrac{1}{4y^2}$ (vi) $\dfrac{1}{x^2} + \dfrac{2}{xy} + \dfrac{1}{y^2}$
Q2: Evaluate Squares Using the Identity
Using the same identity $(a + b)^2 = a^2 + 2ab + b^2$, find the values of the following:
(i) $(64)^2$ (ii) $(105)^2$ (iii) $(205)^2$
(i) $(64)^2$ (ii) $(105)^2$ (iii) $(205)^2$
Strategy: Write each number as a sum of two convenient numbers, then apply $(a + b)^2 = a^2 + 2ab + b^2$.
(i) $(64)^2$
Write $64 = 60 + 4$, so $a = 60$, $b = 4$. $$64^2 = (60 + 4)^2 = 60^2 + 2(60)(4) + 4^2$$ $$= 3600 + 480 + 16 = 4096$$
Write $64 = 60 + 4$, so $a = 60$, $b = 4$. $$64^2 = (60 + 4)^2 = 60^2 + 2(60)(4) + 4^2$$ $$= 3600 + 480 + 16 = 4096$$
(ii) $(105)^2$
Write $105 = 100 + 5$, so $a = 100$, $b = 5$. $$105^2 = (100 + 5)^2 = 100^2 + 2(100)(5) + 5^2$$ $$= 10000 + 1000 + 25 = 11025$$
Write $105 = 100 + 5$, so $a = 100$, $b = 5$. $$105^2 = (100 + 5)^2 = 100^2 + 2(100)(5) + 5^2$$ $$= 10000 + 1000 + 25 = 11025$$
(iii) $(205)^2$
Write $205 = 200 + 5$, so $a = 200$, $b = 5$. $$205^2 = (200 + 5)^2 = 200^2 + 2(200)(5) + 5^2$$ $$= 40000 + 2000 + 25 = 42025$$
Write $205 = 200 + 5$, so $a = 200$, $b = 5$. $$205^2 = (200 + 5)^2 = 200^2 + 2(200)(5) + 5^2$$ $$= 40000 + 2000 + 25 = 42025$$
(i) $64^2 = 4096$ (ii) $105^2 = 11025$ (iii) $205^2 = 42025$