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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.

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.

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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).

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).