The value of [tex]\frac{u}{v} +\frac{v}{u}[/tex] is [tex]\frac{-17}{21}[/tex].
What is quadratic equation?
A quadratic equation is an algebraic equation of the second degree in x. The quadratic equation in its standard form is[tex]ax^{2} +bx+c=0[/tex], where a and b are the coefficients, x is the variable, and c is the constant term.
What is the sum and product of the roots of the quadratic equation?
If [tex]ax^{2} +bx+c = 0[/tex] be the quadratic equation then
Sum of the roots = [tex]\frac{-b}{a}[/tex]
And,
Product of the roots = [tex]\frac{c}{a}[/tex]
According to the given question.
We have a quadratic equation [tex]3x^{2} +5x+7=0..(i)[/tex]
On comparing the above quadratic equation with standard equation or general equation [tex]ax^{2} +bx+c = 0[/tex].
We get
[tex]a = 3\\b = 5\\and\\c = 7[/tex]
Also, u and v are the solutions of the quadratic equation.
⇒ u and v are the roots of the given quadratic equation.
Since, we know that the sum of the roots of the quadratic equation is [tex]-\frac{b}{a}[/tex].
And product of the roots of the quadratic equation is [tex]\frac{c}{a}[/tex].
Therefore,
[tex]u +v = \frac{-5}{3}[/tex] ...(ii) (sum of the roots)
[tex]uv=\frac{7}{3}[/tex] ....(iii) (product of the roots)
Now,
[tex]\frac{u}{v} +\frac{v}{u} = \frac{u^{2} +v^{2} }{uv} = \frac{(u+v)^{2}-2uv }{uv}[/tex] ([tex](a+b)^{2} =a^{2} +b^{2} +2ab[/tex])
Therefore,
[tex]\frac{u}{v} +\frac{v}{u} =\frac{(\frac{-5}{3} )^{2}-2(\frac{7}{3} ) }{\frac{7}{3} }[/tex] (from (i) and (ii))
⇒ [tex]\frac{u}{v} +\frac{v}{u} =\frac{\frac{25}{9}-\frac{14}{3} }{\frac{7}{3} }[/tex]
⇒ [tex]\frac{u}{v} +\frac{v}{u} = \frac{\frac{25-42}{9} }{\frac{7}{3} }[/tex]
⇒ [tex]\frac{u}{v} +\frac{v}{u} = \frac{\frac{-17}{9} }{\frac{7}{3} }[/tex]
⇒ [tex]\frac{u}{v} +\frac{v}{u} =\frac{-17}{21}[/tex]
Therefore, the value of [tex]\frac{u}{v} +\frac{v}{u}[/tex] is [tex]\frac{-17}{21}[/tex].
Find out more information about sum and product of the roots of the quadratic equation here:
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