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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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