Question

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 1

(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.