Question 825531
The roots of the quadratic equation x²-7x+c = 0
differ by 1. Find c
<pre><font size = 4><b>
let r = one root and r-1 be the other root.

Substitute r for x:

Eq. 1:      r²-7r+c = 0

Substitute (r-1) for x

Eq. 2:     (r-1)²-7(r-1)+c = 0

Simplify Eq. 2:

         (r-1)(r-1)-7r+7+c = 0
            r²-2r+1-7r+7+c = 0
                 r²-9r+8+c = 0

So we have this system of equations:

{{{system(r^2-7r+c = 0, r^2-9r+8+c=0)}}}

Get the 8 off the right side of the second equation by
adding -8 to both sides

{{{system(r^2-7r+c = 0, r^2-9r+c=-8)}}}

Multiply the second equation by -1 and add the two equations term by term:

 r²-7r+c = 0
-r²+9r-c = 8
------------
    2r   = 8
       r = {{{8/2}}}
       r = 4

Substitute that in 

         r²-7r+c = 0
{{{(4)^2}}}{{{""-""}}}{{{7(4)}}}{{{""+""}}}{{{c}}} {{{""=""}}} {{{"0"}}}

     {{{16}}}{{{""-""}}}{{{(28)}}}{{{""+""}}}{{{c}}} {{{""=""}}} {{{"0"}}}

          {{{(-12)}}}{{{""+""}}}{{{c}}} {{{""=""}}} {{{"0"}}}

                   {{{c}}} {{{""=""}}} {{{12}}}

Edwin</pre></b>