Let u and v be the solutions to 3x^2 + 5x + 7 = 0. Find u/v+v/u

Question

Let u and v be the solutions to 3x^2 + 5x + 7 = 0. Find u/v+v/u

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Isabelle 3 weeks 2021-11-07T06:05:08+00:00 1 Answer 0 views 0

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    2021-11-07T06:06:36+00:00

    Answer:   \dfrac{-17}{21}

    Step-by-step explanation:

    Given: u and v be are the solutions of  3x^2+5x+7=0

    Let  ax^2+bx+c=0 is the quadratic equation and u and v are the zeroes/solutions then

    Sum of zeroes;   u+v = \dfrac{-b}{a}

    Product of zeroes; uv= \dfrac{c}{a}

    Comparing  3x^2+5x+7=0  to  ax^2+bx+c=0

    we get a= 3 , b= 5 and c = 7

    u+v = \dfrac{-b}{a} = \dfrac{-5}{3}----(i)

    uv= \dfrac{c}{a} = \dfrac{7}{3}----(ii)

    Now we have to find

    \dfrac{u}{v} +\dfrac{v}{u} =\dfrac{u^2+v^2}{uv} adding and subtracting 2uv in numerator we get

    = \dfrac{u^2+v^2+2uv-2uv}{uv}= \dfrac{(u+v)^2-2uv}{uv}

    Substituting the values from (i) and (ii) we get

    \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}

    Hence, the value of \dfrac{u}{v} +\dfrac{v}{u}   is  \dfrac{-17}{21}

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