Question 707166
You have the right values for a, b and c and that is all you need. You do not need to change the expression.<br>
The sum of the roots of a quadratic is -b/a. So the sum of the roots for your expression will be -(-7)/(4) which simplifies to: 7/4<br>
The product of the roots of a quadratic will be c/a. So the product of the roots to your expression will be (3)/(4) = 3/4<br>
If you didn't know that the sum was -b/2a and/or that the product was c/a then it is not terribly difficult to figure this out on your own. From the Quadratic Formula we know that the roots of a general quadratic equation will be:
{{{x[1] = (-b + sqrt(b^2-4ac))/2a}}} and {{{x[2] = (-b - sqrt(b^2-4ac))/2a}}}
Splitting these into two separate fractions will help with the rest:
{{{x[1] = (-b)/2a + sqrt(b^2-4ac))/2a}}} and {{{x[2] = (-b)/2a - sqrt(b^2-4ac)/2a}}}<br>
Now let's see what the sum looks like. If you look at {{{x[1]}}} and {{{x[2]}}} it should be easy to see that the second fractions will cancel each other out because one is positive and the other is negative. So when we add will will get:
{{{(-b)/2a + ((-b)/2a) = (-2b)/2a = (-b)/a}}}<br>
Now let's see what the product looks like:
{{{((-b)/2a + sqrt(b^2-4ac))/2a)((-b)/2a - sqrt(b^2-4ac)/2a)}}}. If we look at this we should be able to see that this fits the {{{p+q)(p-q)}}} pattern. And from the pattern we know that the result is {{{p^2-q^2}}}. Using this pattern will save us time in multiplying:
{{{((-b)/2a)^2 - (sqrt(b^2-4ac))/2a)^2)}}}
which simplifies as follows:
{{{b^2/4a^2 - (b^2-4ac)/4a^2)}}}
These have the same denominator so we can subtract them:
{{{((b^2) - (b^2-4ac))/4a^2)}}}
Simplifying...
{{{4ac/4a^2)}}}
{{{c/a}}}
(Note: I have a terrible memory. I can never remember these sum and product formulas. I always go through the above to figure them out when I need them (like now)).