1.

For a ≠ b, if x + k is the HCF of x2 + ax + b and x2 + bx + a, then the value of a + b is equal to(a) – 2 (b) – 1 (c) 0 (d) 2

Answer»

(b) -1

Since x + k is the HCF of the given expressions, therefore, x = – k will make each expression zero. 

k2 – ak + b = 0                         ...(i) 

k2 – bk + a = 0                       ...(ii) 

Solving (i) and (ii) by the rule of cross multiplication,

\(\frac{k^2}{-a^2+b^2}=\frac{k}{b-a}=\frac{1}{-b+a}\)

From last two relations, k = \(\frac{b-a}{-(b-a)}=-1\)

∴ \(\frac{k^2}{-a^2+b^2}=\frac{1}{-b+a}⇒\frac{(-1)^2}{-(a^2+b^)}=\frac{1}{-b+a}\)

⇒ \(\frac{1}{(b-a)(b+a)}=\frac{1}{-b+a}\) ⇒ a+b = -1.



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