question_answer

(a) Find the value of the expression \[\left( 81{{x}^{2}}+16{{y}^{2}}-72xy \right),\] when \[x=\text{ }\frac{2}{3}\]and \[y=\text{ }\frac{3}{4}\]

(b) If a = 2 and b = 5, then verify \[{{(a+b)}^{2}}=\text{ }{{a}^{2}}+{{b}^{2\text{ }}}+2ab.\]

Answer:

(a)\[81\text{ }{{x}^{2}}+16{{y}^{2}}-72xy={{\left( 9x \right)}^{2}}+{{\left( 4y \right)}^{2}}-2\times 9x\times 4y\]

\[={{(9x-4y)}^{2}}\]

\[[\because \,{{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}]\]

Now, putting \[x=\frac{2}{3}\]and \[y=\frac{3}{4},\]then

\[={{\left( 9\times \frac{2}{3}-4\times \frac{3}{4} \right)}^{2}}\]

\[={{(6-3)}^{2}}={{3}^{2}}=9\]

(b) Putting a = 2 and b = 5, then

L.H.S \[={{\left( a+b \right)}^{2}}\]

\[={{\left( 2+5 \right)}^{2}}={{7}^{2}}=49~\]

and   R.H.S =\[={{a}^{2}}+{{b}^{2}}+2ab\]

\[={{2}^{2}}+{{5}^{2}}+2\times 2\times 5\]

= 4 + 25 + 20 = 49

Hence, L.H.S = R.H.S =49