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Question
If x + a is a common factor of expressions f(x) = x2 + px + q and g(x) = x2 + mx + n; show that : `a = (n - q)/(m - p)`
Solution
f(x) = x2 + px + q
It is given that (x + a) is a factor of f(x).
∴ f(–a) = 0
`\implies` (–a)2 + p(–a) + q = 0
`\implies` a2 – pa + q = 0
`\implies` a2 = pa – q ...(i)
g(x) = x2 + mx + n
It is given that (x + a) is a factor of g(x).
∴ g(–a) = 0
`\implies` (–a)2 + m(–a) + n = 0
`\implies` a2 – ma + n = 0
`\implies` a2 = ma – n ...(ii)
From (i) and (ii), we get,
pa – q = ma – n
n – q = a(m – p)
`a = (n - q)/(m - p)`
Hence, proved.
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