In this unit we are introducing methods of solving a Quadratic Equation: \(ax^2+bx+c=0\). In the Factoring Things unit, we looked at how to factor to solve these equations. This topic focuses on the Method of Square Roots to solve. There are two equations where this method is applied:
| (1) \(ax^2+c=0\) | No \(bx\) term in the equation. |
| (2) \(\left(rx+sy\right)^2=k\) | An expression squared set equal to a constant. |
The process is to get the variable (or expression) by itself on one side of the equation, then reverse the square by taking the square root of BOTH sides. One key point here, we must allow for two solutions by putting \(\pm\).
Solve \(3x^2-54=0\).
| Recognize that the equation does not have a \(bx\) term. This indicates Method of Square Roots needs to be applied. | |
| Algebra to get \(x^2\) by itself: | \(3x^2-54=0\implies3x^2=54\implies x^2=18\) |
| Take square root BOTH sides: | \(\sqrt{x^2}=\color{red}{\pm}\sqrt{18}\) DON'T FORGET THE \(\color{red}{\pm}\)!!! |
| Simplify: | \(x=\pm\sqrt{9\cdot2}\) |
| \(x=\pm3\sqrt{2}\;\color{green}{\checkmark}\) | |
Why the \(\pm\)? Since squares come in opposite pairs: \((3)^2=9\) and \((-3)^2=9\); when the square root is applied there is no indication of BOTH solutions unless we remember this property and write \(\pm\). DON'T FORGET OR YOU ONLY HAVE ONE SOLUTION!
Solve \((2r+3)^2=8\).
Many students make an error and start to multiply out the left side. Recognize that this equation shows an example of type (2), an expression squared set equal to a constant. Proceed the same way:
| Algebra to get the square by itself: | Already by itself: \((2r+3)^2=8\) |
| Take square root BOTH sides: | \(\implies\sqrt{(2r+3)^2}=\color{red}{\pm}\sqrt{8}\) DON'T FORGET THE \(\color{red}{\pm}\)!!! |
| Solve for the variable: | \(\implies2r+3=\pm\sqrt{4\cdot2}\) |
| \(\implies2r=-3\pm2\sqrt{2}\) | |
| \(\implies r=\Large{\frac{-3\pm\sqrt{2}}{2}}\) | |
| An easier simplified form: | \(\implies r=\Large{\frac{-3}{2}\pm\frac{\sqrt{2}}{2}}\;\color{green}{\checkmark}\) |