<PRE><B>Question 179914</B>
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For problem #1, the answer depends on the set of numbers from which you will allow solutions.  They ALL have solutions over the Complex numbers, but some of them do not have solutions if you are only considering Real numbers.  All quadratic equations have 2 roots, all cubic equations have 3, and all nth degree polynomial equations have n roots.  If n is odd, one of the roots is guaranteed to be a real number because complex roots always occur in conjugate pairs.


To determine which of them have solutions over the Reals, compute the discriminant for each one.  For *[tex \Large ax^2 + bx + c = 0], the discriminant, &lt;i&gt;&amp;#916;&lt;/i&gt;, is given by *[tex \Large \Delta = b^2 - 4ac].


If *[tex \Large \Delta &gt; 0] then the two roots are real and unequal.


If *[tex \Large \Delta = 0] then the two roots are real and equal, sometimes referred to as a single root with a multiplicity of two.


If *[tex \Large \Delta &lt; 0] then there are no real number roots.  The two roots are a conjugate pair of complex numbers of the form *[tex \Large a \pm bi] where &lt;i&gt;i&lt;/i&gt; is the imaginary number defined by *[tex \Large i^2 = -1].


Problem 2:


Work backwards.  When you solve a quadratic by factoring, you reach a point where you have something that looks like *[tex \Large (x - a)(x - b) = 0] and then you apply the Zero Product Rule to conclude that &lt;i&gt;x&lt;/i&gt;=&lt;i&gt;a&lt;/i&gt; or &lt;i&gt;x&lt;/i&gt;=&lt;i&gt;b&lt;/i&gt;.  Beginning with &lt;i&gt;x&lt;/i&gt;=&lt;i&gt;-3&lt;/i&gt; or &lt;i&gt;x&lt;/i&gt;=&lt;i&gt;4&lt;/i&gt;, the factored quadratic must be *[tex \Large (x + 3)(x - 4) = 0].  Now all you need to do is multiply the two binomials using FOIL to get your desired quadratic equation.


Problem 3:  I already answered this in my response to Problem 1.  Just pick one of the examples in Problem 1 where *[tex \Large \Delta &lt; 0] and then use the quadratic formula,


*[tex \Large \text{         } x_{1,2} = \frac {-b \pm sqrt{b^2-4ac}}{2a}], 


to find the solutions.  Hint:  For the negative radicand, change the sign to plus and put an &lt;i&gt;i&lt;/i&gt; in front of the radical.  That&#39;s because *[tex \Large sqrt{-p} = sqrt{(-1)p} = i sqrt{p}\text { }(\math \forall p &gt; 0)] 


Problem 4:  John is one year older than his twin brothers, Tom and Jim.  The product of Tom and Jim&#39;s ages is 16.  How old is John?


John is &lt;i&gt;x&lt;/i&gt; years old, so each of the twins is &lt;i&gt;x&lt;/i&gt;-1 years old.  The product of their ages is then *[tex \Large (x - 1)(x - 1) = 16] so:


*[tex \Large x^2 - 2x + 1 = 16]


*[tex \Large x^2 - 2x - 15 = 0]


*[tex \Large (x + 3)(x - 5) = 0]


Throw out the &lt;i&gt;x&lt;/i&gt;= -3 answer because no one can be -3 years old, and you are left with John is 5 years old.


Actually, my answer to 4 is sort of lame.  Maybe you can think of something better.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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