The solution of #ax^2+bx+c=0# is given by #x=(-b+-sqrt(b^2-4ac))/(2a)#
However, if equation is divided by #a#, it becomes of the form #x^2+bx+c=0#, whose solution is #x=(-b+-sqrt(b^2-4c))/2#
and sum of roots is #-b# and product is #c#,
and in #x^2+4x-32=0#, product of roots would be #32# and as its sign is negative, difference of roots is #4#, with smaller root being positive and larger root being negative.
We can follow the following geometric construction, if in #x^2+bx+c=0#, #c<0# and signs of roots are opposite. If #b<0# smaller root is negative (and larger root is positive) and if #b>0#, larger root is negative ((and larger root is negative)).
In such cases, we draw a circle of diameter equal to difference of roots (i.e. #b#) and then a tangent #CP# equal to square-root of product of roots (i.e. #sqrtc#) and then join #P# to the center of circle as shown, then #BP# and #DP# represent the two roots of the equation .
So when draw a circle of diameter #4# units and draw a tangent of length equal to #sqrt32# and join as shown in figure, we will find #BP# and #DP# as #4# and #8# respectively and hence roots are #4# and #-8#.
If in #x^2+bx+c=0#, #c>0#, we can follow the following geometric construction. In this case, if #b<0# both roots are positive and if #b>0#, both roots are negative.
As in such cases, sum of roots and product of roots are known, one can draw a circle with diameter equal to sum of the roots i.e. #b#, (#AB# in figure below) and then draw a line parallel to its diameter at a distance equal to square-root of product of roots i.e. #sqrtc#, (#CD# in figure below with #EF=sqrtc#), cutting circle at #E#. Now draw a perpendicular #EF_|_AB#. Then #AF# and #BF# represent the two roots.
In case any of these construction is not possible, we may have determinant #b^2-4ac<0# and roots are complex.
