Bunuel wrote:
Im2bz2p345 wrote:
The answer will be [C], but I believe it will hold for only odd values of z.
The above expression can be reduced to P = 128 + {[Q^2 -16Q + 64]/2}^z *(-1)^z
=> 128+ [Q-8/2]^2z * (-1)^z
Could someone walk me through ALL the steps in reducing this equation like above? I tried several times to factor it like a quadratic, but I'm way off in my calculations.
Thanks in advance,
~ Im2bz2p345![:)]( "Smile")
The above expression can be reduced to P = 128 + {[Q^2 -16Q + 64]/2}^z *(-1)^z
=> 128+ [Q-8/2]^2z * (-1)^z
Could someone walk me through ALL the steps in reducing this equation like above? I tried several times to factor it like a quadratic, but I'm way off in my calculations.
Thanks in advance,
~ Im2bz2p345
P = 128 + (-\frac{Q^2}{4} + 4Q - 16)^z.
P = 128 + (\frac{-Q^2+16Q-64}{4})^z
P = 128 + (\frac{-(Q-8)^2}{4})^z
P = 128 + (\frac{-(Q-8)^2}{2^2})^z
P = 128 + (-(\frac{Q-8}{2})^2)^z
P = 128 + (-1*(\frac{Q-8}{2})^2)^z
P = 128 + (-1)^z*(\frac{Q-8}{2})^{2z}
I understand all the steps, but what is the logic behind the answer being 8 if the entire solution is dependent upon z, and we are not given any indication of what the value of z could be even or odd or integer??
Thanks