# If the HCF of x²+x-12 and 2x² -kx-9 is x-k , find the value of 'k'.

2

2015-02-17T23:25:35+05:30

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HCF of x²+x-12 and 2x² -kx-9 is x-k so (x-k) is a factor of both the polynomials. So x=k is a zero of both the polynomials. So f(k)=0 in both cases. thus

k² + k - 12 = 0    --------------(1)
2k² - k² - 9 = 0    --------------(2)

from equation 1, k = 3,-4
from equation 2, k = +3, -3
+3 is common to both the solutions.
So value of k is 3.

2015-02-18T12:19:16+05:30

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Let Q(x) = 2 x² - k x - 9
As (x - k) is a factor of  Q(x),   then Q(k) = 0
Hence,  2 k² - k * k - 9 = k² - 9 = 0    =>   k = +3 or -3

As (x - k) is also factor of P(x) = x² + x - 12,   P(k) = 0.

P(3) = 3² + 3 - 12 = 0

Hence  (x - 3)  is the factor of  P(x) and Q(x).