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प्रश्न
If (x – 6) is the HCF of x2 – 2x – 24 and x2 – kx – 6 then the value of k is
पर्याय
3
5
6
8
उत्तर
5
Explanation;
Hint:
HCF = x – 6
p(x) = x2 – 2x – 24
= (x – 6) (x + 4)
g(x) = x2 – kx – 6
∴ x – 6 is the common factor.
g(6) = 62 – k(6) – 6
= 36 – 6k – 6
= 30 – 6k
g(6) = 0
30 – 6k = 0
30 = 6k ⇒ k = `30/6` = 5
The value of k = 5
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