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Consider the equation $(x-p)(x-6)+1=0$ having integral coefficients. If the equation has integral roots, then what values can p have?
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Verified Answer
The correct answer is:
4 or 8
Given equation is $(x-p)(x-6)+1=0$ $\Rightarrow x^{2}-6 x-p x+6 p+1=0$
$\Rightarrow x^{2}-(p+6) x+(6 p+1)=0$
Now, $b^{2}-4 a c=0$
$a=\mathrm{L}, b=-(p+6), c=6 p+1$
$\Rightarrow(p+6)^{2}-4(6 p+1)=0$
$\Rightarrow p^{2}+36+12 p-24 p-4=0$
$\Rightarrow p^{2}-12 p+32=0$
$\Rightarrow(p-4)(p-8)=0$
$\Rightarrow p=4,8$
Hence, $p$ can have 4 or 8 .
$\Rightarrow x^{2}-(p+6) x+(6 p+1)=0$
Now, $b^{2}-4 a c=0$
$a=\mathrm{L}, b=-(p+6), c=6 p+1$
$\Rightarrow(p+6)^{2}-4(6 p+1)=0$
$\Rightarrow p^{2}+36+12 p-24 p-4=0$
$\Rightarrow p^{2}-12 p+32=0$
$\Rightarrow(p-4)(p-8)=0$
$\Rightarrow p=4,8$
Hence, $p$ can have 4 or 8 .
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