If 5 p²-7 p-3=0 and 5 q²-7 q-3=0, p ≠ qthen the equation whose roots are 5 p-4 q and 5 q-4 p is

Correct Answer is : 5x2−7x−439=05 x²-7 x-439=05x2−7x−439=0

Correct Answer is : 5x2−7x−439=05 x²-7 x-439=05x2−7x−439=0

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UPSEE 2017 Solved Paper

Question : 108 of 150

Marks: +1, -0

5 p^{2}-7 p-3=0

5 q^{2}-7 q-3=0, p \neq q

5 p-4 q

5 q-4 p

Solution:

Given,

5 p^{2}-7 p-3=0

.......(i)

5 q^{2}-7 q-3=0

........(ii)

It is clear from Eqs. (i) and (ii)

p

and

q

satisfy the equation

5 x^{2}-7 x-3=0

\therefore p+q=\frac{7}{5}

and

p q=\frac{-3}{5}

........(iii)

We have to find the equation whose roots are

(5 p-4 q)

and

(5 q-4 p)

Here, sum of roots

=(5 p-4 q)+(5 q-4 p)

=p+q

=\frac{7}{5}

[from Eq. (iii)]

And product of roots

=(5 p-4 q)(5 q-4 p)

=81 p q-20(p+q)^{2}

=81 \times \left(\frac{-3}{5}\right)-20 \times \left(\frac{7}{5}\right)^{2}

=\frac{-439}{5}

\therefore

Required equation will be

x^{2}-(

sum of roots)

x+

product of roots

=0

\therefore x^{2}-\frac{7}{5} x+\left(-\frac{439}{5}\right)=0

\Rightarrow 5 x^{2}-7 x-439=0

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