Question

for what values of k,the HCF of x2+x-(2k+2) and 2x2+kx-12 is x+4​

Answer 1

Answer:

k = 5

Step-by-step explanation:

Given :

HCF of {x² + x - (2k + 2) , 2x² + kx - 12} is (x + 4)

Find k,.

Solution :

HCF of {x² + x - (2k + 2) , 2x² + kx - 12} is (x + 4)

⇒ (x + 4) is a factor of both the equations,.

⇒ when x² + x - (2k + 2) , is divided by (x + 4), it gives 0 , as remainder

⇒ If, x + 4 = 0

⇒ x = -4

⇒ x² + x - (2k + 2) = 0

⇒ (-4)² + (-4) - (2k + 2) = 0

⇒ 16 - 4 - 2k - 2 = 0

⇒ 10 - 2k = 0

⇒ 2k = 10 ⇒ k = 5,.

One of the value of k is 5,.

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⇒ (x + 4) is a factor of both the equations,.

⇒ when 2x² + kx - 12 , is divided by (x + 4), it gives 0 , as remainder

⇒ If, x + 4 = 0

⇒ x = -4

⇒ 2x² + kx - 12 = 0

⇒ 2(-4)² + K(-4) - 12 = 0

⇒ 2(16) - 4k - 12 = 0

⇒ 32 - 4k - 12 = 0 ⇒ 20 - 4k = 0

⇒ 4k = 20 ⇒ k = 5,.

∴ The only value of k is 5,.

Answer 2

Answer:

k = 5

Step-by-step explanation

Given : HCF of {x² + x - (2k + 2) , 2x² + kx - 12} is (x + 4)

Find k,.

Solution :

HCF of {x² + x - (2k + 2) , 2x² + kx - 12} is (x + 4)

⇒ (x + 4) is a factor of both the equations,.

⇒ when x² + x - (2k + 2) , is divided by (x + 4), it gives 0 , as remainder

⇒ If, x + 4 = 0

⇒ x = -4

⇒ x² + x - (2k + 2) = 0

⇒ (-4)² + (-4) - (2k + 2) = 0

⇒ 16 - 4 - 2k - 2 = 0

⇒ 10 - 2k = 0

⇒ 2k = 10 ⇒ k = 5,.

One of the value of k is 5,.

__

⇒ (x + 4) is a factor of both the equations,.

⇒ when 2x² + kx - 12 , is divided by (x + 4), it gives 0 , as remainder

⇒ If, x + 4 = 0

⇒ x = -4

⇒ 2x² + kx - 12 = 0

⇒ 2(-4)² + K(-4) - 12 = 0

⇒ 2(16) - 4k - 12 = 0

⇒ 32 - 4k - 12 = 0 ⇒ 20 - 4k = 0

⇒ 4k = 20 ⇒ k = 5,.

∴ The only value of k is 5,.