Exercise 1
Using the identity , expand the following:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
Squaring a sum. The identity \(\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}\) tells you to write down three things: the square of the first term, twice the product of the two terms, and the square of the second term. The only work in each part is deciding which quantity is playing the role of \(\displaystyle a\) and which is playing the role of \(\displaystyle b\).(i) \(\displaystyle (7x+4y)^{2}\), with \(\displaystyle a=7x\) and \(\displaystyle b=4y\):
\[a^{2}=(7x)^{2}=49x^{2},\qquad 2ab=2(7x)(4y)=56xy,\qquad b^{2}=(4y)^{2}=16y^{2}. \]
\[(7x+4y)^{2}=49x^{2}+56xy+16y^{2} \](ii) \(\displaystyle \left(\frac{7}{5}x+\frac{3}{2}y\right)^{2}\), with \(\displaystyle a=\frac{7}{5}x\) and \(\displaystyle b=\frac{3}{2}y\):
\[a^{2}=\frac{49}{25}x^{2},\qquad 2ab=2\cdot\frac{7}{5}\cdot\frac{3}{2}\,xy=\frac{21}{5}xy,\qquad b^{2}=\frac{9}{4}y^{2}. \]
\[\left(\tfrac{7}{5}x+\tfrac{3}{2}y\right)^{2}=\frac{49}{25}x^{2}+\frac{21}{5}xy+\frac{9}{4}y^{2} \](iii) \(\displaystyle (2.5p+1.5q)^{2}\), with \(\displaystyle a=2.5p\) and \(\displaystyle b=1.5q\):
\[a^{2}=6.25p^{2},\qquad 2ab=2(2.5)(1.5)pq=7.5pq,\qquad b^{2}=2.25q^{2}. \]
\[(2.5p+1.5q)^{2}=6.25p^{2}+7.5pq+2.25q^{2} \](iv) \(\displaystyle \left(\frac{3}{4}s+8t\right)^{2}\), with \(\displaystyle a=\frac{3}{4}s\) and \(\displaystyle b=8t\):
\[a^{2}=\frac{9}{16}s^{2},\qquad 2ab=2\cdot\frac{3}{4}\cdot 8\,st=12st,\qquad b^{2}=64t^{2}. \]
\[\left(\tfrac{3}{4}s+8t\right)^{2}=\frac{9}{16}s^{2}+12st+64t^{2} \](v) \(\displaystyle \left(x+\frac{1}{2y}\right)^{2}\), with \(\displaystyle a=x\) and \(\displaystyle b=\frac{1}{2y}\):
\[a^{2}=x^{2},\qquad 2ab=2\cdot x\cdot\frac{1}{2y}=\frac{x}{y},\qquad b^{2}=\frac{1}{4y^{2}}. \]
\[\left(x+\tfrac{1}{2y}\right)^{2}=x^{2}+\frac{x}{y}+\frac{1}{4y^{2}} \](vi) \(\displaystyle \left(\frac{1}{x}+\frac{1}{y}\right)^{2}\), with \(\displaystyle a=\frac{1}{x}\) and \(\displaystyle b=\frac{1}{y}\):
\[a^{2}=\frac{1}{x^{2}},\qquad 2ab=\frac{2}{xy},\qquad b^{2}=\frac{1}{y^{2}}. \]
\[\left(\tfrac{1}{x}+\tfrac{1}{y}\right)^{2}=\frac{1}{x^{2}}+\frac{2}{xy}+\frac{1}{y^{2}} \]Notice that the identity never cared whether \(\displaystyle a\) and \(\displaystyle b\) were whole numbers, fractions, decimals or reciprocals. That is exactly what makes an identity useful: it is true for every value of \(\displaystyle a\) and \(\displaystyle b\).Answers: (i) \(\displaystyle 49x^{2}+56xy+16y^{2}\); (ii) \(\displaystyle \frac{49}{25}x^{2}+\frac{21}{5}xy+\frac{9}{4}y^{2}\); (iii) \(\displaystyle 6.25p^{2}+7.5pq+2.25q^{2}\); (iv) \(\displaystyle \frac{9}{16}s^{2}+12st+64t^{2}\); (v) \(\displaystyle x^{2}+\frac{x}{y}+\frac{1}{4y^{2}}\); (vi) \(\displaystyle \frac{1}{x^{2}}+\frac{2}{xy}+\frac{1}{y^{2}}\).