prove that the equation (b^2 −4ac)x^2 + 4(a + c)x −4 = 0 is always real.


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Question Number 69778 by Rio Michael last updated on 27/Sep/19

$p r o v e t h a t t h e e q u a t i o n$ $(b^{2} - 4 a c) x^{2} + 4 (a + c) x - 4 = 0 i s a l w a y s r e a l .$
Commented by prakash jain last updated on 28/Sep/19

$Hi Rasheed$ $So glad to see that you are still here .$
Commented by prakash jain last updated on 27/Sep/19

$Δ = {(4 (a + c))}^{2} - 4 \times (- 4) (b^{2} - 4 a c)$ $= 16 a^{2} + 16 c^{2} + 32 a c + 16 b^{2} - 64 a c$ $= 16 a^{2} + 16 c^{2} - 32 a c + 16 b^{2}$ $= 16 {(a - c)}^{2} + 16 b^{2}$ ${(a - c)}^{2} is always ⩾ 0$ $b^{2} is always ⩾ 0$ $hence$ $Δ ⩾ 0$ $\Rightarrow roots are always real$ $◼$
Commented by Abdo msup. last updated on 28/Sep/19

$h e l l o p r a k a s h y o u a r e a b s e n t f o r a l o n g t i m e . . .$
Commented by Rasheed.Sindhi last updated on 28/Sep/19

$W e l c o m e s i r p r a k a s h . H a p p y t o s e e$ $y o u i n t h e f o r u m! T h e f o r u m n e e d s y o u .$
Commented by Rio Michael last updated on 28/Sep/19

$t h a n k s s i r$
	Answered by mind is power last updated on 27/Sep/19

	$a = b = c = 0 f a l s e$
	Commented by Rio Michael last updated on 27/Sep/19

	$w h a t d o y o u m e a n s i r ?$
	Commented by mind is power last updated on 27/Sep/19

	$y o u r q u a t i o n m e a n {a x}^{2} + b x + c = 0 . . E l e t y r o o t o f E$ $\Rightarrow (b^{2} - 4 a c) y^{2} + 4 (a + c) y - 4 = 0 ?$
	Commented by JDamian last updated on 28/Sep/19

	$I n t h a t c a s e, t h e r e i s n o e q u a t i o n a t a l l .$
	Answered by MJS last updated on 27/Sep/19

	$usually when dealing with$ ${a x}^{2} + b x + c = 0$ $it is clearly implied that a \neq 0$ $\Rightarrow$ $x^{2} + \frac{4 (a + c)}{b^{2} - 4 a c} x - \frac{4}{b^{2} - 4 a c} = 0; a, b, c \in R \land b^{2} - 4 a c \neq 0$ $x^{2} + p x + q = 0 \Leftrightarrow x = - \frac{p}{2} \pm \sqrt{\frac{p^{2}}{4} - q}$ $this is real for D = \frac{p^{2}}{4} - q ⩾ 0$ $\Rightarrow D = 4 \frac{a^{2} - 2 a c + b^{2} + c^{2}}{{(b^{2} - 4 a c)}^{2}} = 4 \frac{{(a - c)}^{2} + b^{2}}{{(b^{2} - 4 a c)}^{2}} > 0 \Rightarrow$ $\Rightarrow x \in R$
	Commented by Rio Michael last updated on 27/Sep/19

	$t h a n k y o u s i r$
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