Quadratic equations

Worked Out Example 1

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

$\mathbf{ax^{2}-84x+36}$

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

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

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

$\mathbf{\left ( m+n \right )^{2}=m^{2}+2mn+n^{2}}$

Therefore

$\mathbf{ax^{2}-84x+36=m^{2}+2mn+n^{2}}$

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

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

Equate first terms

$\mathbf{{\color{Red} ax^{2}}={\color{Red} m^{2}}}$

Equate second terms

$\mathbf{-{\color{Magenta} 84x}={\color{Magenta} 2mn}}$

Equate third terms

$\mathbf{{\color{Blue} 36}={\color{Blue} n^{2}}}$

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

$\mathbf{-{\color{Magenta} 42x}={\color{Magenta} mn}}$

Square both sides. It rewrites to

$\mathbf{\left ( -1 \right )^{2}{\color{Magenta} \left ( 42x \right )^{2}}={\color{Magenta} \left ( mn \right )^{2}}}$

Which simplifies to

$\mathbf{{\color{Magenta} 42^2x^2}={\color{Magenta} m^2n^2}}$

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

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

From the above two equations, it becomes apparent that