x-2-x-4-x-2-x-4-9-find-the-equation-has-multiple-roots-

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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