Sum of 2 nos.is 8 and their product is 13.11. Find the nos.

• 2.3, 5.7
  
  Let the numbers be x and (8-x), then x(8-x)=13.11
  => x² -8x + 13.11 = 0
  Equation is of the form ax² + bx + c = 0
  Root are given by [-b ± √(b² - 4ac)] / 2a = [8 ± √(64-52.44)]/2 = (8 ± √11.56 )/2
  Hence roots are (8+3.4)/2=5.7 and (8-3.4)/2=2.3
• 6 years agoHelpfull: Yes(1) No(0)
• As. sum is 8, average/midpoint is 4, from which the 2 nos will be equidistant by an unknown factor + or -d .Hence let the 2 nos.be 4+d and 4-d .Sum is 4+d+4-d = 8.Product is (4+d) *(4-d) =16-dsq.=13.11(Given).
  dsq.=16-13.11 =2.89. d=root of 2.89 = 1.7. Hence nos are 4+1.7 and 4-1.7 i.e, 5.7 and 2.3. Here I have not used quadratic equation ,its roots etc.Pl.put this solution in your site for the benefits of your students.In general ,by the same method it can easily be proved that if sum of 2 nos is S and product is P then the nos are :(S/2) +or-sq.root of((Ssq./4)-P))
• 6 years agoHelpfull: Yes(1) No(0)