## Worked Out Example 1

If the following expression is a perfect square, find the value of a

$\fn_jvn&space;\large&space;\mathbf{ax^{2}-84x+36}$

A perfect square expression can be transformed into complete square and vice versa. The above expression is written in complete square form. Now, change it into a perfect square as it is demonstrated here

Suppose you let

$\fn_jvn&space;\large&space;\fn_jvn&space;\large&space;\mathbf{ax^{2}-84x+36}=\mathbf{\left&space;(&space;m+n&space;\right&space;)^{2}}$

Remember the m plus n squared is a perfect square. A perfect square is a contracted form of a complete square.

Expand the RHS to get

$\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;m+n&space;\right&space;)^{2}=m^{2}+2mn+n^{2}}$

Therefore

$\fn_jvn&space;\large&space;\mathbf{ax^{2}-84x+36=m^{2}+2mn+n^{2}}$

As you can see elements on the LHS equals elements on the RHS. For clarity purposes equate same terms on both sides.

$\fn_jvn&space;\large&space;\mathbf{{\color{Red}&space;ax^{2}}-{\color{Magenta}&space;84x}+{\color{Blue}&space;36}={\color{Red}&space;m^{2}}+{\color{Magenta}&space;2mn}+{\color{Blue}&space;n^{2}}}$

Equate first terms

$\fn_jvn&space;\large&space;\mathbf{{\color{Red}&space;ax^{2}}={\color{Red}&space;m^{2}}}$

Equate second terms

$\fn_jvn&space;\large&space;\mathbf{-{\color{Magenta}&space;84x}={\color{Magenta}&space;2mn}}$

Equate third terms

$\fn_jvn&space;\large&space;\mathbf{{\color{Blue}&space;36}={\color{Blue}&space;n^{2}}}$

By observation, it is much easier to start solving from the second term because it is simpler. It comprises of first degree polynomial.

By dividing both sides by 2 the equations reduces to

$\fn_jvn&space;\large&space;\mathbf{-{\color{Magenta}&space;42x}={\color{Magenta}&space;mn}}$

Square both sides. It rewrites to

$\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;-1&space;\right&space;)^{2}{\color{Magenta}&space;\left&space;(&space;42x&space;\right&space;)^{2}}={\color{Magenta}&space;\left&space;(&space;mn&space;\right&space;)^{2}}}$

Which simplifies to

$\fn_jvn&space;\large&space;\mathbf{{\color{Magenta}&space;42^2x^2}={\color{Magenta}&space;m^2n^2}}$

As for the RHS, this is what you need for substitution

$\fn_jvn&space;\large&space;\begin{array}{lcl}&space;\mathbf{{\color{Red}&space;ax^{2}}}&&space;=&space;&&space;\mathbf{{\color{Red}&space;m^{2}}}&space;\\&space;\mathbf{{\color{Blue}&space;36}}&space;&&space;=&space;&&space;\mathbf{{\color{Blue}&space;n^{2}}}&space;\end{array}\$

From the above two equations, it becomes apparent that

$\fn_jvn&space;\large&space;\mathbf{42^{2}x^{2}=36ax^{2}}$

Make a the subject

$\fn_jvn&space;\large&space;\mathbf{a=\frac{42^{2}x^{2}}{36x^{2}}}$

Express the RHS as a product of its prime factors i.e factorize the numerator and denominator to prepare the equation for simplification

$\fn_jvn&space;\large&space;\mathbf{a=\frac{{7}\times&space;{7}\times&space;{6}\times&space;{6}\times&space;x^2}{{6}\times&space;{6}\times&space;x^2}}$

Now simplify by cancelling

$\fn_jvn&space;\large&space;\mathbf{a=\frac{{7}\times&space;{7}\times&space;\cancel{6}\times&space;\cancel{6}\times&space;\cancel{x^2}}{\cancel{6}\times&space;\cancel{6}\times&space;\cancel{x^2}}}$

$\fn_jvn&space;\large&space;\mathbf{a={7}\times&space;{7}}$

$\fn_jvn&space;\large&space;\mathbf{a=49}$