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As zÂ² - 4z - 5 is the kind of a*x^2 + b*x + c where a is positive, thus (z-1)(z+5) > 0 is possible when z is not between the 2 roots : -1 and 5.
Hence, we know that z < -1 or z > 5
The question is a bit strange.... we cannot be sure here ... But as we have to choose, I take (E)
I get: z^2-4z-5>0 (z-5) (z+1)>0 z>5,z>-1
I do not understand the change in sign part. Pls explain.
We apply this rule :
> f(x) = a*x^2 + b*x + c where a is positive, f(x) > 0 when x is not between its roots if these roots exist.
> f(x) = a*x^2 + b*x + c where a is negative, f(x) > 0 when x is between its roots if these roots exist.
If we represent f(z)=z^2-4z-5 in XY plan, we obtain the fig1.
I also add u the representation of g(z)=-z^2-4z+5 to have the case with a <0 in the fig2.
Fig1_a-postive.jpg [ 17.01 KiB | Viewed 530 times ]
Fig2_a-negative.jpg [ 16.88 KiB | Viewed 529 times ]
I got E for this one. My question is why is it this question an odd one ?? examining other choices, only x<-1 is one of the two solution. the other solution which is x>5 is not there and x> -5 or x>-1 range is too wide to make x always true.