Prove the following for the given quadratic

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If the difference of the roots of $x^2-px+q=0$ is unity then prove that $p^2-4q=1$ and $p^2 =4 q^2=({1+2q})^2$
What I Tried
1.I proved the first part of the question using the understanding of the difference of the roots $$D^{1/2}/|a|=|a-b|$$ for $ax^2+bx+c=0$ and D=Discrimininant
2. Do not know where to start.

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There are 2 best solutions below

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You cannot prove it because second part is wrong. For example, the roots of $x^{2}-x=0$ are $0$ and $1$ so the hypothesis is satisfied. We do have $p^{2}-4q=1$ but we do not have $p^{2}=4q^{2}=(1+2q)^{2}$.

Actually, the difference between the roots is $1$ if and only if $p^{2}-4q=1$. Perhaps there is a typo in the question and the first equality in $p^{2}=4q^{2}=(1+2q)^{2}$ is supposed to be replaced by $+$.

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The second one is very likely supposed to be $$ p^2+4q^2 = (1+2q)^2. $$ To prove it, try to expand $(1+2q)^2$ and then use the first result. You no longer need to think about the original quadratic for this part.