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Let p, q and r be real numbers (p ≠ q, r ≠ 0), such that the roots of the
equation \frac{1}{x+p}+\frac{1}{x+q}= \frac{1}{r}  are equal in magnitude but opposite in sign, then the
sum of squares of these roots is equal to :

  • Option 1)

    \frac{p^{2}+q^{2}}{2}

  • Option 2)

    {p^{2}+q^{2}}

  • Option 3)

    2({p^{2}+q^{2}})

  • Option 4)

    {p^{2}+q^{2}+ r^{2}}

 

Answers (1)

best_answer

 

\alpha ^{2}+\beta ^{2}=(\alpha +\beta )^{2}-2\alpha \beta

also \alpha +\beta =0

\alpha ^{2}+\beta^{2} = -2\alpha \beta

r(x+q)+(x+p)= (x+p)(x+q)

x^{2}+x(p+q-2r)+pq- r(p+q)=0

\alpha +\beta =\frac{p+q-2r}{1}=0

p+q = 2r

\left ( \alpha +\beta \right )^{2} = -2[pq-r(p+q)]

\left ( \alpha +\beta \right )^{2} = -2[pq]+(p+q)^{2}]

= p^{2}+q^{2}

 

 

 

 


Option 1)

\frac{p^{2}+q^{2}}{2}

This is incorrect

Option 2)

{p^{2}+q^{2}}

This is correct

Option 3)

2({p^{2}+q^{2}})

This is incorrect

Option 4)

{p^{2}+q^{2}+ r^{2}}

This is incorrect

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