Let f (x) be a quadratic equation of the form:
Where b and c are real numbers.
So a generic procedure for completing squares is:
1) identify b
In this case b = 10
2) divide b between 2 and then square it
3) Add and subtract in the equation
4) Rewrite the equation as follows
The number that you must add to both sides of the equation is number
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Explanation:
The largest rectangle (composed of the green and yellow sections combined) has area of 11*12 = 132 cm^2.
The yellow region takes up 112 of those 132 sq cm. This must mean the green region takes up 132-112 = 20 cm^2.
The horizontal portion of the green rectangle is 12-x cm. The vertical portion is 11-x cm. We can form the area of the green rectangle as an algebraic expression like so
area = length*width
area = (11-x)*(12-x)
area = 132 - 11x - 12x + x^2 .... apply the FOIL rule
area = x^2 - 23x + 132
Set this equal to the 20 cm^2 we found earlier.
x^2 - 23x + 132 = 20
x^2 - 23x + 132-20 = 0
x^2 - 23x + 112 = 0
We could factor or we could use the quadratic formula. I'll go with the second option.
We'll plug in a = 1, b = -23, c = 112
One of these solutions isn't feasible. Note how if x = 16, then this exceeds both the 11 cm and 12 cm sides. So this x value is not possible.
However, x = 7 is possible.
If x = 7, then the horizontal portion of the green rectangle is 12-x = 12-7 = 5 cm. Also, the vertical portion of the green rectangle would be 11-x = 11-7 = 4 cm. The area then is length*width = 5*4 = 20 cm^2 which matches up with what we got earlier. So the answer is confirmed.