Does there exist a pair of consecutive integers whose reciprocals sum to 3/4? Explain how you know.

Accepted Solution

Answer:NOStep-by-step explanation:Let the consecutive numbers be x and x + 1Given condition is their reciprocals sum is equal to 3/4[tex]\frac{1}{x} +\frac{1}{x+1} =\frac{3}{4}[/tex][tex]\frac{2x+1}{x^2+x}=\frac{3}{4}[/tex][tex]8x+4=3x^2+3x[/tex][tex]3x^2-5x-4=0[/tex]Find the discriminant for the above quadratic equationFoe a given quadratic equation [tex]ax^2+bx+c=0[/tex] the discriminant = [tex]b^2-4ac[/tex]Here a = 3,b = -5,c = -4Discriminant = [tex](-5)^2-4\times 3\times (-4)[/tex] =78 >0If a quadratic equation has discriminant > 0 it has distinct real roots The roots are [tex]\frac{-b+\sqrt{b^2-4ac} }{2a}[/tex] and Β  [tex]\frac{-b-\sqrt{b^2-4ac} }{2a}[/tex]Here the roots are [tex]\frac{5+\sqrt{78} }{6} and \frac{5-\sqrt{78} }{6}[/tex]They are not integers So there are no consecutive pair of integers which follow the given condition