# Khan/khan-exercises

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 600eebb New exercise: Completing the square 1 smenks13 authored Nov 4, 2011 1 2 3 4e4cb9b lint: tabs->spaces and jQuery->$for exercises beneater authored Apr 10, 2012 4 5 Completing the square 1 6 7 600eebb New exercise: Completing the square 1 smenks13 authored Nov 4, 2011 12 13 4e4cb9b lint: tabs->spaces and jQuery->$ for exercises beneater authored Apr 10, 2012 14
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18 randRangeNonZero( -10, 10 ) 19 randRange( -5, 5 ) * 2 + ( X1 % 2 ) 20 ( X1 + X2 ) * -1 21 X1 * X2 22 new Polynomial( 0, 2, [C, B, 1], "x" ) 23 POLY.text() 24
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Solve for x by completing the square.

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POLY_TEXT = 0

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Begin by moving the constant term to the right side of the equation.

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x^2 + Bx = C * -1

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We complete the square by taking half of the coefficient of our x term, squaring it, and adding it to both sides of the equation. Since the coefficient of our x term is B, half of it would be B / 2, and squaring it gives us \color{blue}{pow( B / 2, 2 )}.

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x^2 + Bx \color{blue}{ + pow( B / 2, 2 )} = C * -1 \color{blue}{ + pow( B / 2, 2 )}

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We can now rewrite the left side of the equation as a squared term.

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( x + B / 2 )^2 = C * -1 + pow( B / 2, 2 )

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The left side of the equation is already a perfect square trinomial. The coefficient of our x term is B, half of it is B / 2, and squaring it gives us \color{blue}{pow( B / 2, 2 )}, our constant term.

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Thus, we can rewrite the left side of the equation as a squared term.

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( x + B / 2 )^2 = C * -1 + pow( B / 2, 2 )

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Take the square root of both sides.

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x + B / 2 = \pmsqrt( C * -1 + pow( B / 2, 2 ) )

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Isolate x to find the solution(s).

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x = -B / 2\pmsqrt( C * -1 + pow( B / 2, 2 ) )

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x = -B / 2 + sqrt( C * -1 + pow( B / 2, 2 ) ) \text{ or } x = -B / 2 - sqrt( C * -1 + pow( B / 2, 2 ) )

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x = -B / 2 + sqrt( C * -1 + pow( B / 2, 2 ) ) \text{ or } x = -B / 2 - sqrt( C * -1 + pow( B / 2, 2 ) )

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600eebb New exercise: Completing the square 1 smenks13 authored Nov 4, 2011 80 c2cb4db Whitespace :) spicyj authored Dec 8, 2011 81
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