Completing The Square- parts
1+2
The Perfect Square
A perfect square is when a quadratic equation factorises into two brackets that are the same.
For example:
HINT:
x2+8x+16
Notice how (in terms of the formula ax 2+bx+c) ‘b’ is
x2+4x+4x+16 double the number in the brackets and ‘c’ is the square
of the number in the brackets.
x(x+4) +4(x+4)
(x+4) (x+4) or (x+4)2
Completing The Square
But what happens if the quadratic equation isn’t in the form of a perfect square? Can we still use this
method to solve it? Yes, we can.
The steps are relatively similar but contain a few more, such as:
x2
+6x+17 - It is not In terms of ax2+bx+c
the perfect square
1) As shown in the example above, ‘b’ is double the missing number needed
(x+__)2+17 to go into bracket- half of 6 is 3.
(x+3)2+17-__ 2) The perfect square of this would be x 2+6x+9 however because ‘c’ is 17 we
need to take that 9 from 17 (rather than thinking of the perfect square it is
(x+3)2+17-9 easier to square the number in the bracket; 3 2 is 9).
(x+3)2+8 3) Complete the square by taking 9 from 17 to get your answer of +8, making
your completed answer (x+3)2+8.
This method can be used to solve a quadratic equation as well.
What
The usual about negative?
quadratic equation results
x2+8x+29=0 x2-4x+13 in a ‘U’ shape on a graph. However,
(x+4)2 +29-16=0 (x-2)2+13-4 if the equation is something such as
8+2x-x2 the graph will present an ‘n’
(x+4)2 =13 (x-2)2 =9 shape on it. This is because -x2
x+4 =±√13 x-2 =±√9 determines this.
x =-4±√13 x =2±√9
What about when there is a number in front of the x 2?
2x2=12x+37
2(x2+6x) +37
1+2
The Perfect Square
A perfect square is when a quadratic equation factorises into two brackets that are the same.
For example:
HINT:
x2+8x+16
Notice how (in terms of the formula ax 2+bx+c) ‘b’ is
x2+4x+4x+16 double the number in the brackets and ‘c’ is the square
of the number in the brackets.
x(x+4) +4(x+4)
(x+4) (x+4) or (x+4)2
Completing The Square
But what happens if the quadratic equation isn’t in the form of a perfect square? Can we still use this
method to solve it? Yes, we can.
The steps are relatively similar but contain a few more, such as:
x2
+6x+17 - It is not In terms of ax2+bx+c
the perfect square
1) As shown in the example above, ‘b’ is double the missing number needed
(x+__)2+17 to go into bracket- half of 6 is 3.
(x+3)2+17-__ 2) The perfect square of this would be x 2+6x+9 however because ‘c’ is 17 we
need to take that 9 from 17 (rather than thinking of the perfect square it is
(x+3)2+17-9 easier to square the number in the bracket; 3 2 is 9).
(x+3)2+8 3) Complete the square by taking 9 from 17 to get your answer of +8, making
your completed answer (x+3)2+8.
This method can be used to solve a quadratic equation as well.
What
The usual about negative?
quadratic equation results
x2+8x+29=0 x2-4x+13 in a ‘U’ shape on a graph. However,
(x+4)2 +29-16=0 (x-2)2+13-4 if the equation is something such as
8+2x-x2 the graph will present an ‘n’
(x+4)2 =13 (x-2)2 =9 shape on it. This is because -x2
x+4 =±√13 x-2 =±√9 determines this.
x =-4±√13 x =2±√9
What about when there is a number in front of the x 2?
2x2=12x+37
2(x2+6x) +37
