Theory of equations

Question

Suppose $\mathrm{a}\in $I and the equation (x-a) (x-5) = 3 has integral roots, then the set of values which ‘a ’ can take is:

Moderate

Solution

Let m $\in $I be a root of (x - a) (x-5) = 3.then

(m-a) (m -5) = 3

As m-a and m - 5 are integers, $\mathrm{m}-5=\pm 1\text{or}\mathrm{m}-5=\pm 3$

$\Rightarrow $m= 2, 4, 6, 8.

Thus, a = 3,7.

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