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Let $\alpha$ and $\beta$ be the roots of a quadratic equation $4x^2-(5p+1)x+5=0$.If $\beta=1+\alpha,$then find the integral value of $p.$


Sum of roots$=\alpha+\beta=\frac{5p+1}{4}$
Given $\beta=1+\alpha,$so $\alpha-\beta=-1$
Adding the two equations,
$$\alpha=\frac{5p-3}{8}$$
As $\alpha$ is a root of the equation $4x^2-(5p+1)x+5=0$.So $4\alpha^2-(5p+1)\alpha+5=0$
$$4(\frac{5p-3}{8})^2-(5p+1)(\frac{5p-3}{8})+5=0$$
$$5p^2+2p-19=0$$
But when i solve this equation,i do not get integer values of $p.$
But the answer given is that integer value of $p$ is $3.$ I do not know where i am wrong?

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    $\begingroup$ There is nothing wrong with your solution. The problem statement/answer is wrong. If $p=3$, then the equation is $4x^2-16x+5=0$, which has the roots $\alpha=2-\frac{\sqrt{11}}{2}$, $\beta=2+\frac{\sqrt{11}}{2}$, but $|\beta-\alpha|=\sqrt{11}\neq 1$. $\endgroup$ – user236182 Jan 1 '16 at 10:05
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    $\begingroup$ The problem is to prove that the integer part of $p$ is $3$, that is, that $3\leqslant p<4$, not that $p=3$. $\endgroup$ – Did Jan 1 '16 at 10:34
  • $\begingroup$ @Did The roots of $5p^2+2p-19=0$ are approximately $-2.1596$, $1.7596$, so the problem statement/answer is still wrong. $\endgroup$ – user236182 Jan 1 '16 at 10:41
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    $\begingroup$ @user236182 I know--but not for the reasons mentioned on the page when I posted my comment, for example not for the reason in your comment. $\endgroup$ – Did Jan 1 '16 at 10:42

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