Completing the square is an algebraic process that rewrites a quadratic function from standard form into vertex form: $$f(x)=ax^2+bx+c\implies f(x)=a(x-h)^2+k$$ where the point \((h,k)\) represents the vertex. Since this unit focuses on solving equations, the examples show Completing the Square in that context.
Consider \(x^2-6x+8=0\). The steps below outline the Completing the Square process:
| Step 1: Take \(b\) and divide by 2: | \(b=-6\implies\large{\frac{-6}{2}}=-3\) |
| Step 2: Square the result from Step 1: | \(\left(-3\right)^2=9\) |
| Step 3: Now |
\(\left(x^2-6x\color{green}{+9}\right)\color{red}{-9}+8=0\) |
| Step 4: Factor the "new" trinomial in parenthesis from Step 3: | \(\begin{array}{l} (x-3)(x-3)-1=0 \\ \implies(x-3)^2-1=0 \end{array}\) |
| Step 5: Use the Method of Square Roots to finish solving: | \((x-3)^2=1\) |
| \(\implies x-3=\pm\sqrt{1}\) | |
| \(\begin{array}{l} x-3=-1\implies x=2\;\color{green}{\checkmark} \\ x-3=1\implies x=4\;\color{green}{\checkmark} \end{array}\) |
The quadratic equation has to be adjusted to fit in to the examples above. The key point is to find the greatest common factor, GCF, of the first two terms, then complete the square on the factor on the inside.
Consider \(4x^2+32x-40=0\). Use the step by step from above to complete the square:
| Step 1: Factor out the GCF of 1st two terms | \(4\left(x^2+8x\right)-40=0\) |
| Step 2: Now \(b=8\) so divide by \(2\): | \(\frac{8}{2}=4\) |
| Step 3: Square the result from Step 2: | \((4)^2=16\) |
| Step 4: Now |
\(4(x^2+8x\color{green}{+16}\color{red}{-16})-40=0\) |
| Step 5: Factor the trinomial INSIDE the parenthesis: | \(\begin{array}{l} 4\left(\underset{\color{orange}{\text{INSIDE}}}{\color{orange}{\underbrace{x^2+8x+16}}}-16\right)-40=0 \\ \implies4\left(\color{orange}{(x+4)^2}-16\right)-40=0 \end{array}\) |
| Step 6: Distribute through the \(a=\color{blue}{4}\) and combine like terms: | \(\begin{array}{l} \color{blue}{4}(x+4)^2-\color{blue}{4}\cdot(-16)-40=0 \\ \implies4(x+4)^2+64-40=0 \\ \implies4(x+4)^2+24=0 \end{array}\) |
| Now equation is in Method of Square Roots form, finish solving: | \(\begin{array}{l} 4(x+4)^2=-24 \\ \implies(x+4)^2=-6 \\ \implies x+4=\pm\sqrt{-6} \\ \implies x=-4\pm\sqrt{-6} \end{array}\) |
| Right now this equation would have NO SOLUTION, since no real number is formed by the square root of a negative number. To Be Continued . . . | |
Try a final example from start to finish. Solve \(-3x^2+30x+121=0\) by Completing the Square.
Answer: \(x=5\pm\large\frac{\sqrt{196}}{3}\)