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# Khan/khan-exercises

New exercise: Completing the square 1

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1 parent b18ce49 commit 600eebb145dd9a24fb9fc649ca5bf551f3fe08bb smenks13 committed with marcia Nov 4, 2011
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+ randRangeNonZero( -10, 10 ) + randRange( -5, 5 ) * 2 + ( X1 % 2 ) + ( X1 + X2 ) * -1 + X1 * X2 + new Polynomial( 0, 2, [C, B, 1], "x" ) + POLY.text() +
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Solve for x by completing the square.

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POLY_TEXT = 0

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X1
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X2
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x = \quad \quad \text{or} \quad

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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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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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+ X1 +
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x = \quadX1

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