A-nice-question-lt-3-If-a-quadratic-equation-1-q-p-2-2-x-2-p-1-q-x-q-q-1-p-2-2-0-has-equal-roots-prove-that-p-2-4q-

Question Number 113684 by ZiYangLee last updated on 14/Sep/20

A nice question < 3

If a quadratic equation

(1 - q + \frac{p^{2}}{2}) x^{2} + p (1 + q) x + q (q - 1) + \frac{p^{2}}{2} = 0

has equal roots, prove that p^{2} = 4 q

Commented by Dwaipayan Shikari last updated on 14/Sep/20

p^{2} = 4 q

(1 - q + \frac{4 q}{2}) x^{2} + p (1 + q) + q^{2} - q + 2 q = 0

(1 + q) x^{2} + p (1 + q) + q (1 + q) = 0

I t h a s r o o t s

α + β = - p \frac{1 + q}{1 + q} = - p

α β = q = \frac{p^{2}}{4}

(α - β) = \sqrt{p^{2} - 4 . \frac{p^{2}}{4}} = 0

α = β (W h i c h i s t r u e)

Answered by MJS_new last updated on 14/Sep/20

(1 - q + \frac{p^{2}}{2}) x^{2} + p (1 + q) x + q (q - 1) + \frac{p^{2}}{2} = 0

x^{2} + \frac{2 p (q + 1)}{p^{2} - 2 (q - 1)} x + \frac{p^{2} + 2 q (q - 1)}{p^{2} - 2 (q - 1)} = 0

x = t - \frac{p (q + 1)}{p^{2} - 2 (q - 1)}

t^{2} - \frac{(p^{2} - 4 q) (p^{2} + q^{2} - 2 q + 1)}{{(p^{2} - 2 (q - 1))}^{2}} = 0

x_{1} = x_{2} \Leftrightarrow t = 0

\Rightarrow p^{2} - 4 q = 0 \lor p^{2} + q^{2} - 2 q + 1 = 0

\Rightarrow p^{2} = 4 q

(p^{2} + q^{2} - 2 q + 1 = 0 \Rightarrow p^{2} = - {(q - 1)}^{2} \Rightarrow p \notin R)

Commented by ZiYangLee last updated on 15/Sep/20

Cool!