Question

2√2k*(2)²-√2*(2)=0

find k

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Answer 1

Answer:

If x=√2

p(x)=k(x*x) + √2x-4

put x=√2

p(√2)=k*(√2*√2) + √2*(√2)-4

= 2k +(2-4)

= 2k -2

as x=√2 is a solution, so remainder will be equal to 0

therefore, remainder = 2k -2

so,

2k -2 =0

2k= 2

therefore, k = 2/2 = 1

Step-by-step explanation: