Question 100208
<font size = 6><b>Nate's solution is incorrect.</font>
<pre><font size = 4><b>
Let &#945; and &#946; be the two roots of a quadratic equation
ax^2 + 2b^x + c = 0, where a, b and c are constants. 
Obtain the quadratic equation whose two roots are &#955; and &#948;
such that &#955; = &#945;^2 + &#946;^2 and &#948; = &#945;^2 - &#946;^2.

Make use of this

(x - r<sub>1</sub>)(x - r<sub>2</sub>) = 0 has roots r<sub>1</sub> and r<sub>2</sub>

and if you multiply that out by FOIL

x² - r<sub>1</sub>x - r<sub>2</sub>x + r<sub>1</sub>r<sub>2</sub> = 0

and factor x out of the middle two terms:

x² - (r<sub>1</sub> + r<sub>2</sub>)x + r<sub>1</sub>r<sub>2</sub> = 0

This quadratic equations has roots r<sub>1</sub> and r<sub>2</sub>

Since we want the roots to be &#955; and &#948;, we set

r<sub>1</sub> = &#955;, r<sub>2</sub> = &#948;. Substituting:

x² - (&#955; + &#948;)x + &#955;&#948; = 0

Since &#955; = &#945;² + &#946; and &#948; = &#945;² - &#946;².

&#955; + &#948; = (&#945;² + &#946;²) + (&#945;² - &#946;²) = &#945;² + &#946;² + &#945;² - &#946;² = 2&#945;²

&#955;&#948; = (&#945;² + &#946;²)(&#945;² - &#946;²) = &#945;<font size = 2><sup>4</sup></font> - &#946;<font size = 2><sup>4</sup></font>

So

x² - (&#955; + &#948;)x + &#955;&#948; = 0

becomes

x² - 2&#945;²x + (&#945;<font size = 2><sup>4</sup></font> - &#946;<font size = 2><sup>4</sup></font>) = 0

Edwin</pre>