LCM and HCF Practice Test 2

Question : 24 of 40

Marks: +1, -0

If (x - 6) is the HCF of

x^2 - 2x - 24

and

x^ 2 - kx - 6

, then what is the value of k?

Solution:

(x - 6) is the HCF of

x ^2 - 2x - 24

and

x^2 - kx - 6

When

x − 6 = 0

x = 6

Put

x = 6

x^2- kx −6 = 0

(6)^2-6k-6=0

36 - 6 = 6 k

6k = 30

∴ k = 5

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