Question

If ( x- 1) is a factor of kx2 -√2ݔ + 1 , then the value of k is​

Answer 1

Step-by-step explanation:

Hello friends!!

Here is your answer :

P(x) = 6x² + kx - 2

g(x) = 2x - 1

Put g(x) = 0

2x - 1 = 0

x = 1/2

Therefore, g(x) is a factor of p(x).

Then, p( 1/2) = 0

6 {( \frac{1}{2} )}^{2} + k( \frac{1}{2} ) - 2 = 06(

2

1

)

2

+k(

2

1

)−2=0

6 \times \frac{1}{4} - 2+ \frac{1}{2} k = 06×

4

1

−2+

2

1

k=0

\frac{3}{2} - 2 + \frac{1}{2} k = 0

2

3

−2+

2

1

k=0

\frac{3 - 4}{2} + \frac{1}{2} k = 0

2

3−4

+

2

1

k=0

\frac{ - 1}{2} + \frac{1}{2} k = 0

2

−1

+

2

1

k=0

\frac{1}{2} k = \frac{1}{2}

2

1

k=

2

1

k = \frac{1}{2} \times 2k=

2

1

×2

k = 1k=1

Hope it helps you... ^_^

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