Quadratic equations: Solving quadratic equations
Solving quadratic equations by completing the square
Earlier, we saw how the quadratic equation #x^2=c# can be solved by taking the square root of both sides of the equation. This same logic can be applied to solve the quadratic equation #\left(x+a\right)^2=c#, where #a# and #c# are numbers.
# \begin{array}{rcl} \left(x+5\right)^2&=&{{9}\over{4}} \\ &&\phantom{x}\blue{\text{the equation to be solved}}\\ x+5 = -\sqrt{{{9}\over{4}}} &\lor& x+5 = \sqrt{{{9}\over{4}}} \\ &&\phantom{x}\blue{\text{took the square root of both sides}}\\ x+5 = -{{3}\over{2}} &\lor& x+5 = {{3}\over{2}} \\ &&\phantom{x}\blue{\text{calculated}}\\ x=-5-{{3}\over{2}} &\lor& x=-5+{{3}\over{2}} \\ &&\phantom{x}\blue{\text{subtracted }5 \text{ from both sides}}\\ x=-{{13}\over{2}} &\lor& x=-{{7}\over{2}} \\ &&\phantom{x}\blue{\text{calculated}} \end {array} #
Now that we have seen how the quadratic equation #\left(x+a\right)^2=c# can be solved, it would be nice if we were able to rewrite other quadratic expressions in this way. It turns out we can do this by using a method called completing the square.
\[x^2+\green bx+\purple c=\left(x+\frac{\green b}{2}\right)^2-\left(\frac{\green b}{2}\right)^2+\purple c\]
example
# \begin{array}{rcl}
x^2+\green 2x&=&\left(x+\dfrac{\green 2}{2}\right)^2-\left(\dfrac{\green 2}{2}\right)^2\\
&=& (x+1)^2-1
\end {array} #
We can now combine the skill of solving quadratic equations in the form #\left(x+a\right)^2=c# and the method of completing the square to create a general solution for quadratic equations.
Solving a quadratic equation by completing the square
Procedure |
example | |
Solving a quadratic equation by completing the square. |
#2x^2+6x+4=-6x+2# | |
Step 1 |
Reduce the right-hand side of the equation to #0#. |
#2x^2+12x+2=0# |
Step 2 |
Ensure that the coefficient #x^2# equals #1# by dividing both sides by the coefficient of #x^2#. |
#x^2+6x+1=0# |
Step 3 |
Complete the square on the left-hand side. |
#\left(x+3\right)^2-9+1=0# |
Step 4 |
Solve the resulting equation using simplification. |
#x=-3-\sqrt{8} \lor x=-3+\sqrt{8}# |
#\begin{array}{rcl}
x^2-8 x-3&=&0\\
&&\phantom{xxx}\blue{\text{original equation}}\\
(x-4)^2-4^2-3&=&0\\
&&\phantom{xxx}\blue{\text{ }x^2-8x\text{ completed to a square}}\\
(x-4)^2&=&4^2+3\\
&&\phantom{xxx}\blue{\text{everything outside of the brackets moved to the right hand side}}\\
(x-4)^2&=&19\\
&&\phantom{xxx}\blue{\text{simplified the right hand side}}\\
x-4=\sqrt{19} &\lor&x-4=-\sqrt{19}\\
&&\phantom{xxx}\blue{\text{the root taken at both sides}}\\
x=\sqrt{19}+4&\lor& x=4-\sqrt{19}\\
&&\phantom{xxx}\blue{-4 \text{ subtracted from both sides}}\\
\end{array}
#
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