Question 8445
There a several ways to solve this. The easiest is to use the quadratic equation, however I suspect, given the results I get from the quadratic equation, that you are actually required to factor the problem, so, how do you do this?<br>
First, notice that you result will have to have the form<br>
{{{(2x + a)(x + b) = 2x^2 + (a + 2b)x + ab}}}.<br>
The goal is to find the values for a and b. To do this notice that according to the equation,:<br>
a + 2b = 9 and<br>
ab = -5<br>
There are four ways to make ab = -5:<br>
a&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;b&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;ab
&nbsp;5&nbsp;&nbsp;&nbsp;&nbsp;-1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;-5
-5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;-5
&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;-5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;-5
-1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;5&nbsp;&nbsp;&nbsp;&nbsp;-5

One of these combinations must also give us 9, so let's consider what happens with these combinations when we apply them to a + 2b:<br>
a&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;b&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;ab
&nbsp;5&nbsp;&nbsp;&nbsp;&nbsp;-1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;3
-5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;-3
&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;-5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;-9
-1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;5&nbsp;&nbsp;&nbsp;&nbsp;9<br>
The last combination gives us the value we want. This means a = -1 and b = 5 works. So the factorization of {{{2x^2+9x-5=0}}} is {{{(2x - 1)(x + 5)}}}.<br>
This expression is zero when 2x - 1 = 0 or when x + 5 = 0. So the values of x that make the expression equal to zero are x = 1/2 and x = -5.<br>
You can check this by replacing these values in the expression.