Mathematics solution NCERT
Class 9 - Chapter 4: Exploring Algebraic Identities
Question 1. Using the identity $(a+b)^2=a^2+2ab+b^2$, expand the following:
(i) $(7x+4y)^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$Here,
$$ a=7x \qquad\text{and}\qquad b=4y. $$Substitute these values into the identity.
$$ \begin{aligned} (7x+4y)^2 &=(7x)^2+2(7x)(4y)+(4y)^2 \end{aligned} $$Now find each term separately.
$$ \begin{aligned} (7x)^2&=49x^2,\\[4pt] 2(7x)(4y)&=56xy,\\[4pt] (4y)^2&=16y^2. \end{aligned} $$Substituting these values, we get
$$ \begin{aligned} (7x+4y)^2 &=49x^2+56xy+16y^2. \end{aligned} $$Answer:
$$ \boxed{(7x+4y)^2=49x^2+56xy+16y^2} $$(ii) $\left(\dfrac75x+\dfrac32y\right)^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$Here,
$$ a=\frac75x \qquad\text{and}\qquad b=\frac32y. $$Substitute these values into the identity.
$$ \begin{aligned} \left(\frac75x+\frac32y\right)^2 &=\left(\frac75x\right)^2 +2\left(\frac75x\right)\left(\frac32y\right) +\left(\frac32y\right)^2. \end{aligned} $$Now simplify each term.
$$ \begin{aligned} \left(\frac75x\right)^2 &=\frac{49}{25}x^2,\\[6pt] 2\left(\frac75x\right)\left(\frac32y\right) &=\frac{21}{5}xy,\\[6pt] \left(\frac32y\right)^2 &=\frac94y^2. \end{aligned} $$Therefore,
$$ \begin{aligned} \left(\frac75x+\frac32y\right)^2 &=\frac{49}{25}x^2+\frac{21}{5}xy+\frac94y^2. \end{aligned} $$Answer:
$$ \boxed{ \left(\frac75x+\frac32y\right)^2 = \frac{49}{25}x^2+\frac{21}{5}xy+\frac94y^2 } $$(iii) $(2.5p+1.5q)^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$Here,
$$ a=2.5p \qquad\text{and}\qquad b=1.5q. $$Substitute these values into the identity.
$$ \begin{aligned} (2.5p+1.5q)^2 &=(2.5p)^2+2(2.5p)(1.5q)+(1.5q)^2. \end{aligned} $$Now simplify each term.
$$ \begin{aligned} (2.5p)^2 &=6.25p^2,\\[4pt] 2(2.5p)(1.5q) &=7.5pq,\\[4pt] (1.5q)^2 &=2.25q^2. \end{aligned} $$Hence,
$$ \begin{aligned} (2.5p+1.5q)^2 &=6.25p^2+7.5pq+2.25q^2. \end{aligned} $$Answer:
$$ \boxed{(2.5p+1.5q)^2=6.25p^2+7.5pq+2.25q^2} $$Question 1. Using the identity $(a+b)^2=a^2+2ab+b^2$, expand the following:
(iv) $\left(\dfrac34s+8t\right)^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$Here,
$$ a=\frac34s \qquad\text{and}\qquad b=8t. $$Substitute these values into the identity.
$$ \begin{aligned} \left(\frac34s+8t\right)^2 &=\left(\frac34s\right)^2 +2\left(\frac34s\right)(8t) +(8t)^2. \end{aligned} $$Now simplify each term.
$$ \begin{aligned} \left(\frac34s\right)^2 &=\frac{9}{16}s^2,\\[6pt] 2\left(\frac34s\right)(8t) &=12st,\\[6pt] (8t)^2 &=64t^2. \end{aligned} $$Substituting these values, we get
$$ \begin{aligned} \left(\frac34s+8t\right)^2 &=\frac{9}{16}s^2+12st+64t^2. \end{aligned} $$Answer:
$$ \boxed{ \left(\frac34s+8t\right)^2 = \frac{9}{16}s^2+12st+64t^2 } $$(v) $\left(x+\dfrac{1}{2y}\right)^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$Here,
$$ a=x \qquad\text{and}\qquad b=\frac{1}{2y}. $$Substitute these values into the identity.
$$ \begin{aligned} \left(x+\frac{1}{2y}\right)^2 &=x^2 +2\left(x\right)\left(\frac{1}{2y}\right) +\left(\frac{1}{2y}\right)^2. \end{aligned} $$Now simplify each term.
$$ \begin{aligned} x^2 &=x^2,\\[6pt] 2\left(x\right)\left(\frac{1}{2y}\right) &=\frac{xy}{y} =\frac{x}{y},\\[6pt] \left(\frac{1}{2y}\right)^2 &=\frac{1}{4y^2}. \end{aligned} $$Therefore,
$$ \begin{aligned} \left(x+\frac{1}{2y}\right)^2 &=x^2+\frac{x}{y}+\frac{1}{4y^2}. \end{aligned} $$Answer:
$$ \boxed{ \left(x+\frac{1}{2y}\right)^2 = x^2+\frac{x}{y}+\frac{1}{4y^2} } $$(vi) $\left(\dfrac1x+\dfrac1y\right)^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$Here,
$$ a=\frac1x \qquad\text{and}\qquad b=\frac1y. $$Substitute these values into the identity.
$$ \begin{aligned} \left(\frac1x+\frac1y\right)^2 &=\left(\frac1x\right)^2 +2\left(\frac1x\right)\left(\frac1y\right) +\left(\frac1y\right)^2. \end{aligned} $$Now simplify each term.
$$ \begin{aligned} \left(\frac1x\right)^2 &=\frac1{x^2},\\[6pt] 2\left(\frac1x\right)\left(\frac1y\right) &=\frac2{xy},\\[6pt] \left(\frac1y\right)^2 &=\frac1{y^2}. \end{aligned} $$Therefore,
$$ \begin{aligned} \left(\frac1x+\frac1y\right)^2 &=\frac1{x^2}+\frac2{xy}+\frac1{y^2}. \end{aligned} $$Answer:
$$ \boxed{ \left(\frac1x+\frac1y\right)^2 = \frac1{x^2}+\frac2{xy}+\frac1{y^2} } $$Question 2. Using the same identity, find the values of the following:
(i) $(64)^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$Express $64$ as the sum of two convenient numbers.
$$ 64=60+4. $$Here,
$$ a=60 \qquad\text{and}\qquad b=4. $$Using the identity,
$$ \begin{aligned} (64)^2 &=(60+4)^2\\ &=60^2+2(60)(4)+4^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 60^2&=3600,\\[4pt] 2(60)(4)&=480,\\[4pt] 4^2&=16. \end{aligned} $$Add these values.
$$ \begin{aligned} (64)^2 &=3600+480+16\\ &=4096. \end{aligned} $$Answer:
$$ \boxed{64^2=4096} $$(ii) $(105)^2$
Solution:
Express $105$ as
$$ 105=100+5. $$Here,
$$ a=100 \qquad\text{and}\qquad b=5. $$Using the identity,
$$ \begin{aligned} (105)^2 &=(100+5)^2\\ &=100^2+2(100)(5)+5^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 100^2&=10000,\\[4pt] 2(100)(5)&=1000,\\[4pt] 5^2&=25. \end{aligned} $$Add these values.
$$ \begin{aligned} (105)^2 &=10000+1000+25\\ &=11025. \end{aligned} $$Answer:
$$ \boxed{105^2=11025} $$(iii) $(205)^2$
Solution:
Express $205$ as
$$ 205=200+5. $$Here,
$$ a=200 \qquad\text{and}\qquad b=5. $$Using the identity,
$$ \begin{aligned} (205)^2 &=(200+5)^2\\ &=200^2+2(200)(5)+5^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 200^2&=40000,\\[4pt] 2(200)(5)&=2000,\\[4pt] 5^2&=25. \end{aligned} $$Add these values.
$$ \begin{aligned} (205)^2 &=40000+2000+25\\ &=42025. \end{aligned} $$Answer:
$$ \boxed{205^2=42025} $$