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Question Number 120929 by Canovas last updated on 04/Nov/20

Commented by Canovas last updated on 04/Nov/20

$P l e a s e h e l p m e o n n u m b e r 11 a n d 15$
Commented by liberty last updated on 04/Nov/20
it is better if you write
	Answered by Ar Brandon last updated on 04/Nov/20

	$15 .$ ${ax}^{2} + bx + c = 0$ $Let the roots be α and 2 α$ $SumOfRoots : 3 α = - \frac{b}{a} \Rightarrow α = - \frac{b}{3 a}$ $ProductOfRoots : 2 α^{2} = \frac{c}{a}$ $\Rightarrow 2 {(\frac{b}{3 a})}^{2} = \frac{c}{a} \Rightarrow {2 b}^{2} = 9 ac$
	Answered by Ar Brandon last updated on 04/Nov/20

	$11 .$ $x^{2} + 2 px + q = 0$ $Let roots be β and β + 2$ $SumOfRoots : 2 β + 2 = - 2 p \Rightarrow β = - (p + 1)$ $ProductOfRoots : β^{2} + 2 β = q$ $\Rightarrow {(p + 1)}^{2} - 2 (p + 1) = q$ $\Rightarrow (p + 1) (p + 1 - 2) = q$ $\Rightarrow (p + 1) (p - 1) = q$ $\Rightarrow p^{2} - 1 = q, p^{2} = 1 + q$
	Answered by ebi last updated on 04/Nov/20

	$Q 11$ $x^{2} + 2 p x + q = 0$ $t o s h o w p^{2} = 1 + q$ $l e t α a n d β a r e t h e r o o t s o f t h e$ $e q u a t i o n$ $g i v e n t h a t,$ $α - β = 2 \to β = α - 2$ $t h u s,$ $x^{2} - (α + β) x + α β = 0$ $x^{2} - (α + α - 2) x + (α - 2) α = 0$ $x^{2} - (2 α - 2) x + (α^{2} - 2 α) = 0$ $2 α - 2 = - 2 p \to α = 1 - p . . . . . (1)$ $α^{2} - 2 α = q . . . . . . (2)$ $s u b s t i t u t e (1) i n t o (2)$ ${(1 - p)}^{2} - 2 (1 - p) = q$ $1 - 2 p + p^{2} - 2 + 2 p = q$ $p^{2} - 1 = q$ $p^{2} = 1 + q (s h o w n)$
	Answered by ebi last updated on 04/Nov/20

	$Q 15$ ${a x}^{2} + b x + c = 0$ $x^{2} + (\frac{b}{a}) x + (\frac{c}{a}) = 0$ $t o s h o w : 2 b^{2} - 9 a c$ $l e t α a n d β a r e t h e r o o t s o f t h e$ $e q u a t i o n .$ $g i v e n t h a t$ $α = 2 β$ $t h u s,$ $x - (α + β) x + α β = 0$ $x - (2 β + β) x + (2 β) β = 0$ $x - (3 β) x + (2 β^{2}) = 0$ $3 β = - \frac{b}{a} \to β = - \frac{b}{3 a} . . . . . (1)$ $2 β^{2} = \frac{c}{a} . . . . . (2)$ $s u b s t i t u t e (1) i n t o (2)$ $2 {(- \frac{b}{3 a})}^{2} = \frac{c}{a}$ $2 (\frac{b^{2}}{9 a^{2}}) = \frac{c}{a}$ $2 b^{2} = 9 a c \to 2 b^{2} - 9 a c = 0 (s h o w n)$
	Answered by mindispower last updated on 04/Nov/20

	$X^{2} + 2 p X + q = 0 . . E$ $l e t a, b r o o t s o f b - a = 2$ $w e h a v e b + a = - 2 p \Rightarrow p = - (\frac{b + a}{2})$ $a b = q, p^{2} = \frac{1}{4} (b^{2} + a^{2} + 2 a b) = \frac{1}{4} ({(b - a)}^{2} + 4 a b)$ $= \frac{1}{4} (2^{2} + 4 q) = 1 + q = p^{2}$
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