Question

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

Answer 1

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  $\dfrac{-17}{21}$