MathML HTML

Completing the Square

This MathML example demontrates how to complete the square to find the minimum or maximum point of a parabola.

CompleteSquare.html

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  <head>
    <title>XoaX.net's MathML: Completing the Square</title>
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  <body>
    <p>The minimum or maximum value of a parabola can be found by <b>completing the square</b>, as shown. The squared
    	term is greater than or equal to zero. So, it is minimized when it is zero, at the double root. The size of the
    	multiplier <b><i>a</i></b> determines whether it is a minimum or maximum. Finally, the minimum or maximum is
    	given by the value of the last term.</p>
    <math display="block">
    	<mtable>
   			<mtr>
   				<mtd><mi>y</mi></mtd>
    			<mtd><mo>=</mo></mtd>
   				<mtd style="text-align: left;"><mi>a</mi><msup><mi>x</mi><mn>2</mn></msup>
   					<mo>+</mo><mi>b</mi><mi>x</mi><mo>+</mo><mi>c</mi></mtd>
   			</mtr>
   			<mtr>
    			<mtd></mtd>
   				<mtd><mo>=</mo></mtd>
   				<mtd style="text-align: left;"><mi>a</mi><mo>(</mo><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo>
   					<mfrac><mi>b</mi><mi>a</mi></mfrac><mi>x</mi><mo>)</mo><mo>+</mo><mi>c</mi></mtd>
   			</mtr>
   			<mtr>
    			<mtd></mtd>
   				<mtd><mo>=</mo></mtd>
   				<mtd style="text-align: left;"><mi>a</mi><mo>(</mo><msup><mi>x</mi><mn>2</mn></msup>
   					<mo>+</mo>
   					<mfrac><mi>b</mi><mi>a</mi></mfrac><mi>x</mi>
   					<mo>+</mo>
   					<msup><mrow><mo>(</mo><mfrac><mi>b</mi><mrow><mn>2</mn><mi>a</mi></mrow></mfrac><mo>)</mo></mrow><mn>2</mn></msup>
   					<mo>-</mo>
   					<msup><mrow><mo>(</mo><mfrac><mi>b</mi><mrow><mn>2</mn><mi>a</mi></mrow></mfrac><mo>)</mo></mrow><mn>2</mn></msup>
   					<mo>)</mo><mo>+</mo><mi>c</mi></mtd>
   			</mtr>
				<mtr>
    			<mtd></mtd>
   				<mtd><mo>=</mo></mtd>
   				<mtd style="text-align: left;"><mi>a</mi><mo>(</mo>
   					<msup><mrow><mo>(</mo><mi>x</mi><mo>+</mo>
   						<mfrac><mi>b</mi><mrow><mn>2</mn><mi>a</mi></mrow></mfrac>
   						<mo>)</mo></mrow><mn>2</mn></msup>
   					<mo>-</mo>
   					<msup><mrow><mo>(</mo><mfrac><mi>b</mi><mrow><mn>2</mn><mi>a</mi></mrow></mfrac><mo>)</mo></mrow><mn>2</mn></msup>
   					<mo>)</mo><mo>+</mo><mi>c</mi></mtd>
   			</mtr>
   			<mtr>
    			<mtd></mtd>
   				<mtd><mo>=</mo></mtd>
   				<mtd style="text-align: left;"><mi>a</mi>
   					<msup><mrow><mo>(</mo><mi>x</mi><mo>+</mo>
   						<mfrac><mi>b</mi><mrow><mn>2</mn><mi>a</mi></mrow></mfrac>
   						<mo>)</mo></mrow><mn>2</mn></msup>
   					<mo>-</mo>
   					<mfrac><msup><mi>b</mi><mn>2</mn></msup><mrow><mn>4</mn><mi>a</mi></mrow></mfrac>
   					<mo>+</mo><mi>c</mi></mtd>
   			</mtr>
   			<mtr>
    			<mtd></mtd>
   				<mtd><mo>=</mo></mtd>
   				<mtd style="text-align: left;"><mi>a</mi>
   					<msup><mrow><mo>(</mo><mi>x</mi><mo>+</mo>
   						<mfrac><mi>b</mi><mrow><mn>2</mn><mi>a</mi></mrow></mfrac>
   						<mo>)</mo></mrow><mn>2</mn></msup>
   					<mo>+</mo>
   					<mfrac>
   						<mrow><mn>4</mn><mi>a</mi><mi>c</mi><mo>-</mo><msup><mi>b</mi><mn>2</mn></msup></mrow>
   						<mrow><mn>4</mn><mi>a</mi></mrow>
   					</mfrac>
   				</mtd>
   			</mtr>
   		</mtable>
   	</math>
   	<p>To make the situation clearer, we can write
   	<math>
   		<mi>y</mi><mo>=</mo><mi>a</mi><msup><mi>z</mi><mn>2</mn></msup><mo>+</mo><mi>d</mi>
   	</math>,
   	where
   	<math>
   		<mi>z</mi><mo>=</mo><mi>x</mi><mo>+</mo><mfrac><mi>b</mi><mrow><mn>2</mn><mi>a</mi></mrow></mfrac>
   	</math>
   	and
   	<math>
   		<mi>d</mi><mo>=</mo>
   		<mfrac>
   			<mrow><mn>4</mn><mi>a</mi><mi>c</mi><mo>-</mo><msup><mi>b</mi><mn>2</mn></msup></mrow>
   			<mrow><mn>4</mn><mi>a</mi></mrow>
   		</mfrac>
   	</math>.
   	</p>
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