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(ax-7)(bx+5) = 6x2+cx-35    If ab=6 and a-b=-1 what are two possible values of c?

I don't know if I'm on the right track, but so far I've found out that c=5a-7b

Please solve this problem for me and show me how you did it! Thanks :)

a) c=2 or c=-4

b) c=11 or c=-11

c) c=-2 or c=4

d) c=12 or c=-5

Feb 14, 2018

#1
+1

a=2

b=3

(2x-7)(3x+5) substitute

6x²+10x-21x-35 expand

6x²-11x-35 rearrange

c=-11

or

c=11

choose option b)   Feb 14, 2018
edited by lynx7  Feb 14, 2018
edited by lynx7  Feb 14, 2018
edited by lynx7  Feb 14, 2018
#2
+2

(ax-7)(bx+5) = 6x^2+cx-35    If ab = 6 and a - b =  -1 what are two possible values of c?

(ab x^2  - 7bx + 5ax  -  35)  =  6x^2 + cx  -35

6x^2  + ( 5a - 7b) x - 35   =   6x^2  +  cx  -  35

5a  - 7b   =  c

If   a  - b  = - 1      then     b  =  a + 1

And

ab  = 6   ⇒   b   =  6/a

So

b   =   b

a  + 1   =  6/a           multiply through by a    and rearrange

a^2  + a  -  6   =  0   factor

(a - 2)  ( a + 3)  =  0

Set each factor to 0  and solve and we get that   a  = 2  , a  = -3

When a  =  2      b  =  6 / a   =  3

When  a  = - 3     b  =  6/-3  =  -2

So

(a, b )  =   (2, 3)   or  ( -3, -2)

Using   5a  - 7b  = c

5(2)  - 7(3)  =  c  =  - 11

5(-3) - 7(-2)  = c  =  -1

Proof

a = 2  b = 3

( 2x - 7) ( 3x +  5)  =  6x^2  - 21x + 10x - 35   = 6x^2 - 11x - 35   ⇒   c  =  11

a = - 3, b = -2

(-3x - 7) (-2x + 5)  =   6x^2  - 15x + 14x - 25 = 6x^2 - 1x - 35       ⇒  c  = -1   Feb 14, 2018