4F - Completing the square
The process of completing the square (CTS) allows us to convert quadratics in the general form \(y=ax^2+bx+c\) into turning point form \(y=a(x-h)^2+k\). When in turning point form, the turning point (or vertext) is located at \((h,k)\).
Method for completing the square:
Provided the coefficient of \(x^2\) is \(1\), use the following steps to complete the square:
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■ To complete the square the coefficient of \(x^2\) must be \(1\). If the coefficient is not \(1\) then remove it as a common factor from all of the terms: \[ax^2+bx+c=a\left [ x^2+\frac{bx}{a}+\frac{c}{a} \right ] \] |
4F - Example 1: Completing the square
Complete the square on the quadratic \(x^2+6x-2\) and hence state the coordinates of the turning point.
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4F - Example 1: Video solution
4F - Example 1: Practice
Question 1: Express \(y=x^2+6x-15\) in the form \(y=a(x-h)^2+k\) and hence state the coordinates of the turning point. Question 2: Express \(y=x^2-10x+9\) in the form \(y=a(x-h)^2+k\) and hence state the coordinates of the turning point. 4F - Example 1: Solutions
Question 1: \(y=(x+3)^2-24\) Therefore, the turning point is at \((-3,-24)\). Question 2: \(y=(x-5)^2-16\) Therefore, the turning point is at \((5,-16)\). |
4F - Example 2: Completing the square
Complete the square on the quadratic \(x^2+8x-11\) and hence state the coordinates of the turning point.
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4F - Example 2: Video solution
4F - Example 2: Practice
Question 1: Complete the square on the quadratic \(x^2-4x+7\) and hence state the coordinates of the turning point. Question 2: Complete the square on the quadratic \(x^2+12x+24\) and hence state the coordinates of the turning point. Question 3: Complete the square on the quadratic \(x^2+3x+5\) and hence state the coordinates of the turning point. 4F - Example 2: Solutions
Question 1: \(y=(x-2)^2+3\) Therefore, the turning point is at \((2,3)\). Question 2: \(y=(x+6)^2-12\) Therefore, the turning point is at \((-6,-12)\). Question 3: \(y=(x+\frac{3}{2})^2+\frac{11}{4}\) Therefore, the turning point is at \((-\frac{3}{2}, \frac{11}{4})\). |
4F - Example 3: Completing the square for a non-monic trinomial
Complete the square on the quadratic \(4x^2-2x-5\) and hence state the coordinates of the turning point.
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4F - Example 3: Video solution
4F - Example 3: Practice
Question 1: Complete the square on \(y=2x^2+4x-17\) and hence state the coordinates of the turning point. Question 2: Complete the square on \(y=-x^2-8x-7\) and hence state the coordinates of the turning point. Question 3: Complete the square on \(y=4x^2-6x+10\) and hence state the coordinates of the turning point. 4F - Example 3: Solutions
Question 1: \(y=2(x+1)^2-19\) Therefore, the turning point is at \((-1,-19)\). Question 2: \(y=-(x+4)^2+9\) Therefore, the turning point is at \((-4,9)\). Question 3: \(y=4(x-\frac{3}{4})^2+\frac{31}{4}\) Therefore, the turning point is at \((\frac{3}{4},\frac{31}{4})\). |
Solving quadratic equations by completing the square:
Once you have completed the square on a quadratic, it is possible to find the solutions to the equation by using inverse operations to isolate \(x\). This only works because there is 1 term involving \(x\).
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■ Remember that \(\sqrt{a^2}=\pm a\) |
4F - Example 4: Solving quadratic equations by completing the square
Solve the following quadratic for \(x\), given it is already in turning point form: \[(x+4)^2-27=0\]
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4F - Example 4: Video solution
4F - Example 4: Practice
Question 1: Solve the equation \((x-6)^2-16=0\) for \(x\). Question 2: Solve the equation \((x+3)^2-12=0\) for \(x\). Question 3: Solve the equation \(-2(x+5)^2+8=0\) for \(x\). 4F - Example 4: Solutions
Question 1: \[(x-6)^2=16\] \[\therefore x-6=\pm \sqrt{16}\] \[\therefore x=6\pm 4\] \[\therefore x=2, x=10\] Question 2: \[(x+3)^2=12\] \[\therefore x+3=\pm \sqrt{12}\] \[\therefore x=-3\pm 2\sqrt{3}\] \[\therefore x=-3-2\sqrt{3}, x=-3+2\sqrt{3}\] Question 3: \[-2(x+5)^2=-8\] \[\therefore (x+5)^2=4\] \[\therefore x+5=\pm \sqrt{4}\] \[\therefore x=-5\pm 2\] \[\therefore x=-7, x=-3\] |
4F - Example 5: Solving quadratic equations by completing the square
Solve the following quadratic for \(x\), by first completing the square: \[x^2-4x-9=0\]
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4F - Example 5: Video solution
4F - Example 5: Practice
Question 1: Solve the equation \(x^2+6x+16=0\) for \(x\) by first completing the square. Question 2: Solve the equation \(-x^2+8x-11=0\) for \(x\) by first completing the square. 4F - Example 5: Solutions
Question 1: \[(x+3)^2-25=0\] \[\therefore (x+3)^2=25\] \[\therefore x+3=\pm 5\] \[\therefore x=-8, x=2\] Question 2: \[-(x-4)^2+5=0\] \[\therefore (x-4)^2=5\] \[\therefore x-4=\pm \sqrt{5}\] \[\therefore x=4-\sqrt{5}, x=4+\sqrt{5}\] |
The axis of symmetry:
By completing the square on the general quadratic form \(ax^2+bx+c\) we find \[ax^2+bx+c=...=a\left ( x+\frac{b}{2a} \right )-\frac{b^2}{4a^2}+c\] ■ For a quadratic \(y=ax^2+bx+c\), the turning point (or vertex) always exists on the axis of symmetry which is given by \[x=-\frac{b}{2a}\] Consider the parabola \(y=x^2-8x+1\) as shown. The axis of symmetry is \[x=-\frac{-8}{2(1)}=4\]
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4F - Example 6: The axis of symmetry and completing the square
Use the axis of symmetry to determine the turning point for the following quadratic. Hence, determine the turning point form without completing the square. \[y=x^2-3x+5\]
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4F - Example 6: Video solution
4F - Example 6: Practice
Question 1: Determine the coordinates of the turning point of the parabola \(y=x^2-4x-6\). Hence, state the rule of the quadratic in turning point form. Question 2: Determine the coordinates of the turning point of the parabola \(y=2x^2+12x+24\). Hence, state the rule of the quadratic in turning point form. 4F - Example 6: Solutions
Question 1: \[x=-\frac{-4}{2(1)}=2\] \[\therefore y=(2)^2-4(2)-6=-10\] Therefore, the turning point is at \((2,-10)\) \[\therefore y=(x-2)^2-10\] Question 2: \[x=-\frac{12}{2(2)}=-3\] \[\therefore y=2(-3)^2+12(-3)+24=6\] Therefore, the turning point is at \((-3,6)\) \[\therefore y=2(x+3)^2+6\] |
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Additional Exercises
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Topic Worksheets
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Other Resources
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Completing the square - Worksheet A
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