Completing the square
Completing the square in a quadratic expression where the coefficient of x2 is non-unitary
When completing the square, expressions are written in the form:
\(y = a{(x - b)^2} + c\)
Example
Rewrite \(y = {x^2} - 6x + 11\) in the form \(y = a{(x - b)^2} + c\).
In this case \(a = 1\) as the coefficient of \({x^2}\) is \(1\).
To get the number inside the bracket, we half the coefficient of the second term in the original equation, in this case \(6\).
This gives \({(x - 3)^2}\).
If we expand \({(x - 3)^2}\), we get:
\(= (x - 3)(x - 3)\)
\(= {x^2} - 3x - 3x + 9\)
\(= {x^2} - 6x + 9\)
You can see that the \({x^2} - 6x\) matches the first two terms from the original equation. But we have also added on an extra \(9\) from this squared bracket, so we need to take it off again to ensure that we don't change the value of the expression. This gives:
\(y = {(x - 3)^2} - 9 + 11\)
\(y = {(x - 3)^2} + 2\)
Now try these questions.
Question
Rewrite \(y = {x^2} + 10x + 7\) in the form \(y = {(x - b)^2} + c\).
Since \((x + 5)(x + 5) = {x^2} + 5x + 5x + 25\)
\(= {x^2} + 10x + 25\), we need to subtract 25.
\(y = {(x + 5)^2} - 25 + 7\)
\(y = {(x + 5)^2} - 18\)
Question
Rewrite \(y = 5 - 2x - {x^2}\) in the form \(y = a{(x - b)^2} + c\).
First rearrange the equation so that the \({x^2}\) term is first:
\(y = - {x^2} - 2x + 5\)
Next we need to take a common factor of \(- 1\) outside the brackets so that the \({x^2}\) has coefficient 1.
\(y = - 1({x^2} + 2x - 5)\)
Now we can complete the square as before, then multiply out the bracket at the end.
Since \((x + 1)(x + 1) = {x^2} + x + x + 1\)
\(= {x^2} + 2x + 1\), we need to subtract 1.
\(y = - 1\left[ {{{(x + 1)}^2} - 1 - 5} \right]\)
\(y = - 1\left[ {{{(x + 1)}^2} - 6} \right]\)
\(y = - {(x + 1)^2} + 6\)
This can also be written as \(y = 6 - {(x + 1)^2}\)
Question
Extension
Rewrite \(y = 2{x^2} + 6x + 7\) in the form \(y = a{(x - b)^2} + c\)
Firstly, take out a common factor of \(2\), so that we have \({x^2}\) on its own.
\(y = 2\left( {{x^2} + 3x + \frac{7}{2}} \right)\)
\(y = 2\left[ {{{\left( {x + \frac{3}{2}} \right)}^2} - \frac{9}{4} + \frac{7}{2}} \right]\)
Remember...since \(\left( {x + \frac{3}{2}} \right)\left( {x + \frac{3}{2}} \right)\)
\(= {x^2} + \frac{3}{2}x + \frac{3}{2}x + \frac{9}{4}\)
\(= {x^2} + 3x + \frac{9}{4}\), you need to subtract \(\frac{9}{4}\).
So:
\(y = 2\left[ {{{\left( {x + \frac{3}{2}} \right)}^2} + \frac{5}{4}} \right]\)
\(y = 2{\left( {x + \frac{3}{2}} \right)^2} + \frac{5}{2}\)
Once you've completed the square you'll be able to write the co-ordinates and the nature of the turning point.
Therefore the quadratic above has a minimum turning point at \(\left( { - \frac{3}{2},\frac{5}{2}} \right)\).