Question 185592
Let the two positive real numbers be x and y, then...
1) {{{x+y = 7}}} "Two positive real numbers have a sum of 7..."
2) {{{x*y = 11}}} "...and a product of 11."
Rewrite equation 1) as x = 7-y and substitute this for x in equation 2)
2a) {{{(7-y)*y = 11}}} Perform the indicated multiplication (distribute) on the left side.
2a) {{{7y-y^2 = 11}}} Rearrange to form a standard quadratic equation.
2a) {{{y^2-7y+11 = 0}}} Use the quadratic formula: {{{x = (-b+-sqrt(b^2-4ac))/2a}}} to solve (a = 1, b = -7, c = 11):
{{{x = (-(-7)+-sqrt((-7)^2-4(1)(11)))/2(1)}}} Simplify:
{{{x = (7+-sqrt(49-44))/2}}}
{{{x = (7+-sqrt(5))/2}}}
{{{highlight(x = (7+sqrt(5))/2)}}} or {{{highlight(x = (7-sqrt(5))/2)}}}
Check:
{{{((7+sqrt(5))/2)+((7-sqrt(5))/2) = (14+sqrt(5)-sqrt(5))/2}}} = {{{14/2 = 7}}}
{{{((7+sqrt(5))/2)*((7-sqrt(5))/2) = (49-5)/4}}} = {{{44/4 = 11}}}
On part 2 of the problem, your "weird" answer is the correct solution!
{{{highlight(x = (-3/2)+sqrt(3)/2)}}} 
{{{highlight(x = (-3/2)-sqrt(3)/2)}}}