Answer: [tex]\dfrac{-17}{21}[/tex]
Step-by-step explanation:
Given: u and v be are the solutions of [tex]3x^2+5x+7=0[/tex]
Let [tex]ax^2+bx+c=0[/tex] is the quadratic equation and u and v are the zeroes/solutions then
Sum of zeroes; [tex]u+v = \dfrac{-b}{a}[/tex]
Product of zeroes; [tex]uv= \dfrac{c}{a}[/tex]
Comparing [tex]3x^2+5x+7=0[/tex] to [tex]ax^2+bx+c=0[/tex]
we get a= 3 , b= 5 and c = 7
[tex]u+v = \dfrac{-b}{a} = \dfrac{-5}{3}----(i)[/tex]
[tex]uv= \dfrac{c}{a} = \dfrac{7}{3}----(ii)[/tex]
Now we have to find
[tex]\dfrac{u}{v} +\dfrac{v}{u} =\dfrac{u^2+v^2}{uv}[/tex] adding and subtracting 2uv in numerator we get
[tex]= \dfrac{u^2+v^2+2uv-2uv}{uv}= \dfrac{(u+v)^2-2uv}{uv}[/tex]
Substituting the values from (i) and (ii) we get
[tex]\dfrac{(\dfrac{-5}{3} )^2-2\times \dfrac{7}{3} }{\dfrac{7}{3} } = \dfrac{\dfrac{25}{9} -\dfrac{14}{3} }{\dfrac{7}{3} }= \dfrac{\dfrac{25-42}{9} }{\dfrac{7}{3}} =\dfrac{-17}{9} \times \dfrac{3}{7} = \dfrac{-17}{21}[/tex]
Hence, the value of [tex]\dfrac{u}{v} +\dfrac{v}{u}[/tex] is [tex]\dfrac{-17}{21}[/tex]
