If the equation (2x−1)−p(x^2 +2)=0, where p is constant, has imaginary roots, deduce that 2p^2 +p−1≥0.


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Question Number 165651 by MathsFan last updated on 06/Feb/22

$I f t h e e q u a t i o n (2 x - 1) - p (x^{2} + 2) = 0,$ $w h e r e p i s c o n s t a n t, h a s i m a g i n a r y$ $r o o t s, d e d u c e t h a t 2 p^{2} + p - 1 ⩾ 0 .$
Commented by mr W last updated on 06/Feb/22

${p x}^{2} - 2 x + 2 p + 1 = 0$ $x = \frac{2 \pm \sqrt{2^{2} - 4 p (2 p + 1)}}{2 p} = \frac{1 \pm \sqrt{1 - p - 2 p^{2}}}{p}$ $i f 1 - p - 2 p^{2} ⩾ 0 \Rightarrow r e a l r o o t s$ $i f 1 - p - 2 p^{2} < 0 \Rightarrow c o m p l e x r o o t s :$ $x = \frac{1 \pm i \sqrt{2 p^{2} + p - 1}}{p}$ $s i n c e \frac{1}{p} \neq 0, t h e e q u a t i o n c a n n e v e r$ $h a v e i m a g i n a r y r o o t s!$ $t h e r e f o r e : q u e s t i o n i s w r o n g!$ $1) w h e n 2 p^{2} + p - 1 = 0 i t h a s r e a l r o o t s .$ $2) w h e n 2 p^{2} + p - 1 > 0 i t h a s c o m l e x r o o t s .$
Commented by peter frank last updated on 06/Feb/22

$thank you$
Commented by MathsFan last updated on 06/Feb/22

$t h a n k s$
Commented by chrisbridge last updated on 06/Feb/22

$i t h i n k y o u d o n o t n e e d \frac{1}{p} t o b e e q u a l t o 0 t o h a v e i m a g i n a r y r o o t s . Y o u o n l y n e e d t h e d i s c r i m a n t t o b e = 0 . F r o m y o u r w o r k a b o v e 2 p^{2} + p - 1 > 0 i s t r u e f o r t h e f i r s t e q u a t i o n t o h a v e i m a g i n a r y r o o t s$
Commented by mr W last updated on 06/Feb/22

$i {d o n}^{'} t n e e d, b u t t h e q u e s t i o n n e e d s!$ $t h e q u e s t i o n r e q u e s t s i m a g i n a r y r o o t s .$ $b i i s a n i m a g i n a r y n u m b e r .$ $a + b i i s a c o m p l e x n u m b e r, n o t a n$ $i m a g i n a r y n u m b e r, i f a \neq 0 .$ $s i n c e \frac{1}{p} {c a n}^{'} t b e z e r o, s o t h e e q u a t i o n$ $c a n n e v e r h a v e i m a g i n a r y r o o t s!$
	Answered by chrisbridge last updated on 06/Feb/22

	$b^{2} - 4 a c < 0 f o r i m a g i n a r y r o o t s$ $2 x - 1 - {p x}^{2} - 2 p = 0$ ${p x}^{2} - 2 x + (1 + 2 p) = 0$ ${(- 2)}^{2} - 4 p (1 + 2 p) < 0$ $4 - 4 p - 8 p^{2} < 0$ $4 - 4 p - 8 p^{2} < 0$ $d i v i d i n g t h r o u g h b y 4$ $1 - p - 2 p^{2} < 0$ $o n m u l t i p l y i n g t h r o u g h b y (- 1)$ $- 1 + p + 2 p^{2} > 0$ $r e a r r a g i n g$ $2 p^{2} + p - 1 > 0$ $r e q u i r e d s o l u t i o n$
	Commented by mr W last updated on 07/Feb/22

	$y o u s e e m t o t h i n k t h a t i m a g i n a r y$ $n u m b e r s a n d c o m p l e x n u m b e r s a r e$ $t h e s a m e t h i n g .$
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