I got the one but not the -5/7. You have 1 and half minutes average time do these problems do you really think you can use the quadractic equation in that little time? I think Jeff b is a little more on target.

The only way I could figure it out was to realize 1 for x in denominator provided zero( I was lucky to see it) then I knew that 1 less than any of the answers would provide the correct solution when f(x) also equaled zero

since d was less than one, we got a negative value 2/7-1 =-5/7 when that was plugged into the denominator.
(5^2/7^2)*7 = 25/7, -5/7* -2 =+10/7 , -5
it provided ( 25/7 + 10/7) -5 = 0
35/7 = 5, 5 – 5 = 0

I had also factored out the denominator and got the sames as jeff b ( -7x-5) (-x+1) but could not put it together. that was the most efficient way.

## Comments (4)

Why not just factor the denominator into (7X + 5)(X-1)=0; so X=1 and X= -5/7, quicker, no?

I got the one but not the -5/7. You have 1 and half minutes average time do these problems do you really think you can use the quadractic equation in that little time? I think Jeff b is a little more on target.

Why not just use -a/b to find the sum of the denominator?

The only way I could figure it out was to realize 1 for x in denominator provided zero( I was lucky to see it) then I knew that 1 less than any of the answers would provide the correct solution when f(x) also equaled zero

since d was less than one, we got a negative value 2/7-1 =-5/7 when that was plugged into the denominator.

(5^2/7^2)*7 = 25/7, -5/7* -2 =+10/7 , -5

it provided ( 25/7 + 10/7) -5 = 0

35/7 = 5, 5 – 5 = 0

I had also factored out the denominator and got the sames as jeff b ( -7x-5) (-x+1) but could not put it together. that was the most efficient way.