Let-p-and-q-are-the-roots-of-x-2-2mx-5n-0-and-m-and-n-are-the-roots-of-x-2-2px-5q-0-If-p-q-m-n-then-the-value-of-p-q-m-n-is-

Question Number 31320 by Joel578 last updated on 06/Mar/18

Let p and q are the roots of

x^{2} - 2 m x - 5 n = 0

and m and n are the roots of

x^{2} - 2 p x - 5 q = 0

If p \neq q \neq m \neq n, then the value of

p + q + m + n is \dots

Answered by MJS last updated on 06/Mar/18

(x - p) (x - q) = x^{2} - 2 m x - 5 n

x^{2} - (p + q) x + p q = x^{2} - 2 m x - 5 n

p + q = 2 m; p q = - 5 n

m = \frac{p + q}{2}; n = - \frac{p q}{5}

(x - m) (x - n) = x^{2} - 2 p x - 5 q

x^{2} - (m + n) x + m n = x^{2} - 2 p x - 5 q

m + n = 2 p; m n = - 5 q

\frac{p + q}{2} - \frac{p q}{5} = 2 p \Rightarrow p = \frac{5 q}{15 + 2 q}

\frac{p + q}{2} \times \frac{p q}{5} = 5 q \Rightarrow q (p^{2} + p q - 50) = 0 \Rightarrow

\Rightarrow q_{1} = 0; p_{1} = 0 : not valid (p \neq q)

p^{2} + p q - 50 = 0 with p = \frac{5 q}{15 + 2 q}

q^{3} - 10 q^{2} - 300 q - 1225 = 0

try all \pm q with q ∣ 1225

\Rightarrow q_{2} = - 5; p_{2} = - 5; not valid (p \neq q)

q^{2} - 15 q - 225 = 0

q_{3} = \frac{15}{2} (1 - \sqrt{5}); p_{3} = \frac{5}{2} (3 + \sqrt{5})

m_{3} = \frac{5}{2} (3 - \sqrt{5}); n_{3} = \frac{15}{2} (1 + \sqrt{5})

q_{4} = n_{3}; p_{4} = m_{3}

m + n + p + q = 30

Commented by Joel578 last updated on 06/Mar/18

t h a n k y o u v e r y m u c h

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