0
$\begingroup$

Which of the following is NOT true if $p^2 = p + 1$ and p is a real number
A. $p^3 = p^2 + p$
B. $p^4 = p^3 + p + 1$
C. $p^3 = 2p + 1$
D. $ p^3 + p^2 = p + 1$
E. $p = \frac{1 } {p−1}$

I tried factoring it but it did not work.

$\endgroup$
  • $\begingroup$ Hint: just manipulate each expression in turn. For example, $p^3=p^2+p$ follows from $p^2=p+1$ by multiplying through by $p$. And so on. $\endgroup$ – lulu Jan 23 '17 at 11:21
1
$\begingroup$

In D you have $p^3+p^2=p+1$, but $p^2=p+1$ so $$p^3+p^2=p^2$$ $$p^3=0$$ $$p=0$$ but this is not true from $p^2=p+1$

$\endgroup$
1
$\begingroup$

Just manipulate each expression but some cases do follow each other.

As $p^2=p+1$, we have $$p^3=p (p+1) =p^2+p =(p+1) +p =2p+1$$ So $A $ and $C $ are correct.

We have $$p^2-p = p (p-1)=1\Rightarrow p=\frac {1}{p-1} $$, so $E $ follows.

Squaring we get, $$p^4=p^2+2p+1 =(p^2+p)+(p+1) =p^3+p+1$$, so $C $ is correct (from $A $, which is correct, we use $p^3=p^2+p $).

From $C $, we can see clearly that $D $ doesn't follow and is hence the only incorrect option. Hope it helps.

$\endgroup$
  • $\begingroup$ i do not get what u did here: Squaring we get, p4=p2+2p+1=(p2+p)+(p+1)=p3+p+1p4=p2+2p+1=(p2+p)+(p+1)=p3+p+1 I know u swuared (p^2), which is p+1, giving us p^2+2p+1, and p^2=(p+1), however where did the (p^2+p) come from???? $\endgroup$ – exchangehelpforuni Jan 23 '17 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.