
Complete a Square Factoring Help Fun Game Tips:  A perfect square is formed by multiplying a quantity times itself, as in (3)(3) =9 . Here are seven perfect squares: 25, 144, 169, (3)^{2}, r^{2}, (y+3)^{2}, (2w5t)^{2} .  The 'Perfect Trinomial Square' x^{2}+6x+9 can be factored as (x+3)(x+3).  When given only 2 terms such as x^{2}+6x , how can the 3rd term '9' be found to complete the square? The answer is to take half of the '6' to get '3' and square the '3' to get the '9'. When you complete the square for x^{2}+6x you use x^{2}+6x+(6/2)^{2} for the answer x^{2}+6x+9 . A check with x=10, shows 160=(10)^{2}+6(10) changes to the square 169=(10)^{2}+6(10)+(6/2)^{2} by adding '9'.  In general, to complete the square for x^{2}+bx , take half of the b and square it to get the 3rd term as shown here x^{2}+bx+(b/2)^{2} . To complete x^{2}+5x , the answer x^{2}+5x+(2.5)^{2} is the trinomial square x^{2}+5x+6.25 .  When the numerical coefficient of x^{2} is not a simple 'one', for example 2x^{2}+12x , in order to complete the square for 2x^{2}+12x , begin with the common factor '2' to get 2[x^{2}+6x] and then take half of the 6 and square it, to get the trinomial square 2[x^{2}+6x+(6/2)^{2}] = 2[x^{2}+6x+9] .  In general, to complete the square for ax^{2}+bx , use common factoring as in a[x^{2}+(b/a)x] , then take half of b/a and square it, (b/2a)^{2}, to get the perfect trinomial square a[x^{2}+(b/a)x+(b/2a)^{2}] .  Squared quantities are used in formulas for many applications: light and radiation intensity, gravitational attraction, kinetic energy, bending strength, maximum & minimum problems...  Completing the square can be used to find the lowest point [vertex] of a parabola such as y = x^{2}+6x+8. For example, complete the square for y = x^{2}+6x+8 based on squaring half of the '6'. y = x^{2}+6x+99+8. Add '9' and also subtract '9' to keep the y value unchanged. y = (x+3)^{2}9+8. y = (x+3)^{2}1. Since the perfect square (x+3)^{2} is always positive, then the lowest value of 'y' is '1'. The vertex of this sample parabola is (3,1). The same parabola can be represented as y+1 = (x+3)^{2}.  Refresh/Reload the web page to start over again.  If the game doesn't respond to keyboard input, click inside the game area to reset the game's focus.  The score report automatically appears after you have made 10 choices.  The Factors score is based on your choices only and does not count the helicopter hits.  The Game score is reduced by the number of helicopter hits.  The game speeds up as your score increases. Adjust this with the  and + keys. 
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