Question

How do you solve ${{x}^{2}}-2x=32$?

Answer 1

Hint: Graphing, completing the squares, quadratic formula, factoring FOIL, The Diagonal Sum Method, the Bluma Method, the popular factoring AC Method, and the new Transforming Method are the eight most common methods for solving quadratic equations.

Complete step by step answer:
The quadratic formula is the obvious choice when the quadratic equation cannot be factored. Solving by formula, on the other hand, feels tedious and repetitive. In addition, school math curriculum encourages students to learn a few other methods of problem solving in addition to the formula.
Following steps are required to solve quadratic equation using completing square method,
Assume the quadratic equation is \[a{{x}^{2~}}+\text { } bx\ text{ }+\text{ }c~=\text{ }0\]then, using the completing square method, follow the steps to solve it.
Form the equation in such a way that c is on the right side.
If \[a\] is not 1, divide the entire equation by \[a\]to get the co-efficient of \[{{x}^{2}}\] to be 1.
Add the square of half of the term-x co-efficient, \[{{\left (b/2a \right)}^{2}}\] to both sides.
Factor the left side of the equation as the binomial term's square.
On both sides, take the square root.
Find the roots of the variable x by solving for it.
So, let us solve above equation using this steps,
\[x=1-\sqrt{3}\]or \[x=1+\sqrt{3}\]
\[\begin{align}
  & \Rightarrow {{x}^{2}}-2x=32 \\
 & \Rightarrow {{x}^{2}}-2x-32=0 \\
 & \Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
\end{align}\]
\[\Rightarrow x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(1)(-32)}}{2(1)}\]
\[\begin{align}
  & \Rightarrow x=\dfrac{2\pm \sqrt{4+128}}{2} \\
 & \Rightarrow x=\dfrac{2\pm \sqrt{132}}{2} \\
 & \Rightarrow x=\dfrac{2\pm \sqrt{2\cdot 2\cdot 3\cdot 11}}{2} \\
 & \Rightarrow x=\dfrac{2\pm 2\sqrt{33}}{2} \\
 & \Rightarrow x=\dfrac{2-2\sqrt{33}}{2} \\
 & \Rightarrow x=\dfrac{2(1-\sqrt{33})}{2} \\
 & \Rightarrow x=1-\sqrt{33} \\
 & or \\
 & \Rightarrow x=\dfrac{2+2\sqrt{33}}{2} \\
 & \Rightarrow x=1+\sqrt{33} \\
\end{align}\]
Thus, the roots are \[x=1-\sqrt{3}\] or \[x=1+\sqrt{3}\]

Note: Completing the square is advantageous because it provides an alternative to the quadratic formula and can solve problems that the quadratic formula cannot. While the previous problem may have been factored, there is one example where this formula is required.