First slide

Theory of equations

Question

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

Moderate

A

$ϕ$
B

{-11, - 13}
C

{3,7}
D

{-3,-7}

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, $m - 5 = \pm 1 or m - 5 = \pm 3$
$\Rightarrow$ m= 2, 4, 6, 8.
Thus, a = 3,7.

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