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Solve, writing any non-real solutions in the form a+ bi

0 votes

a) (x^2-4x-3)(x^2+9)=0

b)5x^2=2x^3- x^4

asked Oct 2, 2018 in ALGEBRA 2 by anonymous
reshown Oct 2, 2018 by bradely

1 Answer

0 votes

a)

( x^2 - 4x - 3 ) ( x^2 + 9 )  =  0

( x^2 - 4x - 3 )  =  0                                        ;       ( x^2 + 9 )  =  0

x^2 - 4x - 3  =  0                                            ;       x^2  =  - 9

x  = { -(-4) ± √[ (-4)^2 - 4(1)(-3) ] } / 2(1)        ;       x  =  - 9

x  = { 4 ± √[ 16 + 12 ] } / 2                              ;       x  =  - 9

x  = { 4 ± √28 } / 2                                          ;       x  =  ± 3i

x  = { 4 ± 2√7 } / 2                                          ;       x  =  3i       ;   x  =  -3i

x  =  4/2  ± (2√7)/2                                         ;       x  =  3i       ;   x  =  -3i

x  =  2 ± 2√7                                                  ;       x  =  3i       ;   x  =  -3i

x  =  2 + 2√7        ;        x  =  2 + 2√7             ;       x  =  3i       ;   x  =  -3i

 

b)

5x^2  =  2x^3 - x^4

5x^2  =  x^2 ( 2x - x^2)

5  =  2x - x^2

2x^2 - 2x + 5  =  0

2x^2 - 2x + 5  =  0

x  = { -(-2) ± √[ (-2)^2 - 4(2)(5) ] } / 2(2) 

x  = { 2 ± √[ 4 - 40] } / 4

x  = { 2 ± √(-36)} / 4

x  = { 2 ± 6i)} / 4

x  = 2(1 ± 3i) / 4

x  = (1 ± 3i) / 2

x  = (1 + 3i) / 2   ;    x  = (1 - 3i) / 2

Answer :

a)  The Solutions are x  =  2 + 2√7,  x  =  2 + 2√7, x  =  3i and x  =  -3i.

b) The Solutions are x  = (1 + 3i) / 2 and x  = (1 - 3i) / 2.

answered Oct 4, 2018 by homeworkhelp Mentor

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