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Fixed typos (#27202)

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Andrew Mackie
2019-03-07 18:46:51 -08:00
committed by Randell Dawson
parent edb3548826
commit d37f5a7751
1 changed files with 5 additions and 5 deletions
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10
guide/english/mathematics/completing-the-square/index.md
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@ -13,7 +13,7 @@ A quadratic equation generally takes the form: <em>ax<sup>2</sup> + bx + c</em>
<pre><em>ax<sup>2</sup> + bx = -c</em></pre>
2. Make the coefficient of x<sup>2</sup> equal to 1 by dividing both sides of the equation by <em>a</em> so that we now have: <br>
<pre>x<sup>2</sup> + (<sup>b</sup>/<sub>a</sub>)x = - (<sup>c</sup>/<sub>a</sub>)</pre>
<pre>x<sup>2</sup> + (<sup>b</sup>/<sub>a</sub>)x = -(<sup>c</sup>/<sub>a</sub>)</pre>
3. Next, add the square of half of the coefficient of the <em>x</em>-term to both sides of the equation: <br>
<pre>x<sup>2</sup> + (<sup>b</sup>/<sub>a</sub>)x + (<sup>b</sup>/<sub>2a</sub>)<sup>2</sup> = (<sup>b</sup>/<sub>2a</sub>)<sup>2</sup> - (<sup>c</sup>/<sub>a</sub>)</pre>
@ -21,14 +21,14 @@ A quadratic equation generally takes the form: <em>ax<sup>2</sup> + bx + c</em>
4. Completing the square on the Left Hand Side and simplifying the Right Hand Side of the above equation, we have:
<pre>(x<sup></sup> + <sup>b</sup>/<sub>2a</sub>)<sup>2</sup> = (<sup>b<sup>2</sup></sup>/<sub>4a<sup>2</sup></sub>) - (<sup>c</sup>/<sub>a</sub>)</pre>
5. Further simplpfying the Right Hand Side,
<pre>(x<sup></sup> + <sup>b</sup>/<sub>2a</sub>)<sup>2</sup> = (b<sup>2</sup> - 4ac)/4a<sup>2</sup> </pre>
5. Further simplifying the Right Hand Side,
<pre>(x<sup></sup> + <sup>b</sup>/<sub>2a</sub>)<sup>2</sup> = (b<sup>2</sup> - 4ac) &divide; 4a<sup>2</sup> </pre>
6. Finding the square root of both sides of the equation,
<pre>x<sup></sup> + <sup>b</sup>/<sub>2a</sub> = &radic;(b<sup>2</sup> - 4ac) &divide; 2a </pre>
<pre>x<sup></sup> + <sup>b</sup>/<sub>2a</sub> = &pm;((b<sup>2</sup> - 4ac)<sup>&half;</sup> &divide; 2a) </pre>
7. By making x the subject of our formula, we are able to solve for its value completely:
<pre>x<sup></sup> = -b &#177; &radic;(b<sup>2</sup> - 4ac) &divide; 2a </pre>
<pre>x<sup></sup> = (-b &pm; (b<sup>2</sup> - 4ac)<sup>&half;</sup>) &divide; 2a </pre>
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