Question

Express $\dfrac{4}{25}{{m}^{2}}-mn+\dfrac{25}{16}{{n}^{2}}$ as square of a binomial, also evaluate for the given values $m=5$ , $n=8$ .

Answer 1

Hint: In this problem we need to convert the given expression into the square of the binomial and calculate the value of expression for the given values of the variables. We know that the binomial is the expression which contains two terms only in addition or subtraction. So we need to reduce the given expression into two terms which are in addition or subtraction with a square also. So we will simplify the each term and try to convert them in any form like ${{a}^{2}}-2ab+{{b}^{2}}$ or ${{a}^{2}}+2ab+{{b}^{2}}$ so that we can use the algebraic formulas ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ or ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . After applying the algebraic formula we can substitute the given values of the variables and simplify the expression to get the required result.

Complete step by step solution:
Given expression is $\dfrac{4}{25}{{m}^{2}}-mn+\dfrac{25}{16}{{n}^{2}}$.
We have the coefficient of ${{m}^{2}}$ as $\dfrac{4}{25}$ and the coefficient of ${{n}^{2}}$ as $\dfrac{25}{16}$ .
We can write the values $4$ , $16$ and $25$ as ${{2}^{2}}$ , ${{4}^{2}}$ and ${{5}^{2}}$ respectively. Then the given expression will be modified as
$\dfrac{4}{25}{{m}^{2}}-mn+\dfrac{25}{16}{{n}^{2}}=\dfrac{{{2}^{2}}}{{{5}^{2}}}{{m}^{2}}-mn+\dfrac{{{5}^{2}}}{{{4}^{2}}}{{n}^{2}}$
Applying the exponential formula $\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$ , then we will get
$\dfrac{4}{25}{{m}^{2}}-mn+\dfrac{25}{16}{{n}^{2}}={{\left( \dfrac{2}{5} \right)}^{2}}{{m}^{2}}-mn+{{\left( \dfrac{5}{4} \right)}^{2}}{{n}^{2}}$
Applying the exponential formula ${{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m}}$ , then we will have
$\dfrac{4}{25}{{m}^{2}}-mn+\dfrac{25}{16}{{n}^{2}}={{\left( \dfrac{2m}{5} \right)}^{2}}-mn+{{\left( \dfrac{5n}{4} \right)}^{2}}$
Writing the middle term of the above expression as $mn=2\left( \dfrac{2m}{5} \right)\left( \dfrac{5n}{4} \right)$ . Then the above equation is modified as
$\dfrac{4}{25}{{m}^{2}}-mn+\dfrac{25}{16}{{n}^{2}}={{\left( \dfrac{2m}{5} \right)}^{2}}-2\left( \dfrac{2m}{5} \right)\left( \dfrac{5n}{4} \right)+{{\left( \dfrac{5n}{4} \right)}^{2}}$
Applying the algebraic formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ in the above equation, then we will get
$\dfrac{4}{25}{{m}^{2}}-mn+\dfrac{25}{16}{{n}^{2}}={{\left( \dfrac{2m}{5}-\dfrac{5n}{4} \right)}^{2}}$
Hence the value of the given expression $\dfrac{4}{25}{{m}^{2}}-mn+\dfrac{25}{16}{{n}^{2}}$ as square of a binomial is ${{\left( \dfrac{2m}{5}-\dfrac{5n}{4} \right)}^{2}}$ .
Substituting the values $m=5$ , $n=8$ in the above equation, then we will have
$\begin{align}
  & \dfrac{4}{25}{{\left( 5 \right)}^{2}}-\left( 5 \right)\left( 8 \right)+\dfrac{25}{16}{{\left( 8 \right)}^{2}}={{\left( \dfrac{2}{5}\times 5-\dfrac{5}{4}\times 8 \right)}^{2}} \\
 & \Rightarrow \dfrac{4}{25}{{\left( 5 \right)}^{2}}-\left( 5 \right)\left( 8 \right)+\dfrac{25}{16}{{\left( 8 \right)}^{2}}={{\left( 2-10 \right)}^{2}} \\
 & \Rightarrow \dfrac{4}{25}{{\left( 5 \right)}^{2}}-\left( 5 \right)\left( 8 \right)+\dfrac{25}{16}{{\left( 8 \right)}^{2}}={{\left( -8 \right)}^{2}} \\
 & \Rightarrow \dfrac{4}{25}{{\left( 5 \right)}^{2}}-\left( 5 \right)\left( 8 \right)+\dfrac{25}{16}{{\left( 8 \right)}^{2}}=64 \\
\end{align}$
Hence the value of the given expression $\dfrac{4}{25}{{m}^{2}}-mn+\dfrac{25}{16}{{n}^{2}}$ for the values $m=5$ , $n=8$ is $64$ .

Note: In this problem we have used the exponential formulas as well as the algebraic formulas to solve the problem. To solve similar kind of problems some of the exponential and algebraic formulas are given below
${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ , $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ and $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ .