Question

3.Find the value of k, if x - 1 is a factor of p(x) in each of the
if x – 1 is a factor of p(x) in each of the following cases:
(i) P(x) = x2 + x + k
(ii) p(x) = 2x² + kx +root 2
(iii) p(x) = kx2 - root 2x+1
(iv) p(x) = kx2 – 3x +k​

Answer 1

The values of k in each case is as follows:-

case i: k = -2

case ii: k = -(2 + root 2)

case iii: k = root 2 - 1

case iv: k = 3/2

Given:

(1) (x - 1) is a factor of p(x)

(2) P(x) in each of the following cases:

    (i) P(x) = x2 + x + k

    (ii) P(x) = 2x² + kx +root 2

    (iii) P(x) = kx2 - root 2x+1

    (iv) P(x) = kx2 – 3x +k​

To find:

The value of k

Solution:

Since (x - 1) is a factor of P(x), then x - 1 = 0 should satisfy P(x) i.e., for x = 1, P(x) = 0 or in other words, P(1) = 0.

(i)

P(x) = x2 + x + k

=> P(1) = 1^2 + 1 + k = 0

=> 2 + k = 0

=> k = -2

(ii)

P(x) = 2x² + kx +root 2

=> P(1) = 2*1² + k*1 +root 2 = 0

=> 2 + k + root 2 = 0

=> k = - 2 - root 2

=> k = -(2 + root 2)

(iii)

P(x) = kx2 - root 2x + 1

=> P(1) = k*(1)^2 - (root 2)*1 + 1 = 0

=> k - root 2 + 1 = 0

=> k = root 2 - 1

(iv)

P(x) = kx2 – 3x + k​

=> P(1) = k*(1)^2 - 3*1 + k = 0

=> 2k - 3 = 0

=> 2k = 3

=> k = 3/2

The values of k in each case is as follows:-

case i: k = -2

case ii: k = -(2 + root 2)

case iii: k = root 2 - 1

case iv: k = 3/2

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