(x^2 +x−4)(x^2 +x+4)=9 find the equation has multiple roots.


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Question Number 11500 by @ANTARES_VY last updated on 27/Mar/17

$(x^{2} + x - 4) (x^{2} + x + 4) = 9$ $find the equation has multiple roots .$
	Answered by ridwan balatif last updated on 27/Mar/17

	$((x^{2} + x) - 4) ((x^{2} + x) + 4) = 9$ ${(x^{2} + x)}^{2} - 16 = 9$ ${(x^{2} + x)}^{2} - 25 = 0$ $((x^{2} + x) - 5) ((x^{2} + x) + 5) = 0$ $(i) x^{2} + x - 5 = 0$ $D = b^{2} - 4 ac$ $= 1^{2} - 4 (1) (- 5)$ $= 21$ $x_{1, 2} = \frac{- b \pm \sqrt{D}}{2 a}$ $= \frac{- 1 \pm \sqrt{21}}{2}$ $x_{1} = \frac{- 1 + \sqrt{21}}{2}$ $x_{2} = - (\frac{1 + \sqrt{21}}{2})$ $(ii) x^{2} + x + 5 = 0$ $D = b^{2} - 4 ac$ $= 1^{2} - 4 (1) (5)$ $= - 19 (IMPOSSIBLE)$ $point (ii) has no roots$
	Commented by @ANTARES_VY last updated on 27/Mar/17

	$The answer is : - 5 .$
	Commented by @ANTARES_VY last updated on 27/Mar/17

	$error$
	Commented by mrW1 last updated on 27/Mar/17

	$p l e a s e e x p l a i n w h a t y o u m e a n w i t h$ $y o u r q u e s t i o n .$
	Commented by mrW1 last updated on 27/Mar/17

	$t h e n y o u r q u e s t i o n s h o u l d b e : f i n d t h e$ $p r o d u c t o f t h e r e a l r o o t s o f t h e$ $e q u a t i o n .$
	Commented by @ANTARES_VY last updated on 27/Mar/17

	$x_{1} \times x_{2} = ?$
	Commented by mrW1 last updated on 27/Mar/17

	$y e s . x_{1} \times x_{2} = - 5$
	Commented by sandy_suhendra last updated on 27/Mar/17

	$from the answer of Mr Ridwan Balatif$ $x^{2} + x + 5 = 0 has 2 imaginary roots because D < 0$ $x^{2} + x - 5 = 0 has 2 real roots x_{1} and x_{2} because D > 0$ $so x_{1} . x_{2} = \frac{c}{a} = \frac{- 5}{1} = - 5$
	Answered by ajfour last updated on 27/Mar/17

	$x = \frac{- 1 \pm \sqrt{1 \pm 20}}{2} .$
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