Question

plzz guys help me out..........
$2x {}^{2} + kx + \sqrt{2}$
$find \: the \: value \: of \: k \: if \: x - 1is \: a \: \: factor \: of \: px \\ in \: each \: of \: the \: following \: cases$

Answer 1

Given f(x) = 2x^2 + kx + root 2.

Given g(x) = x - 1.

By the Remainder theorem, we get

= > x - 1 = 0

= > x = 1.

Plug x = 1 in f(x), we get

$= \ \textgreater \ 2(1)^2 + k(1) + \sqrt{2} = 0$

$= \ \textgreater \ 2 + k + \sqrt{2} = 0$

$= \ \textgreater \ k = -2 - \sqrt{2}$



Hope this helps!

Answer 2

$\textbf { Hey Mate }$

$\textbf { your answer is -- }$

Given, f(x) = 2x^2+kx+√2

since, x-1 is a factor of given f(x)

so, x-1=0. => x = 1

therefore , f(1) = 0

=> 2×1^2+k×1+√2 = 0

=> 2+k+√2 = 0

=> k = -2-√2

hence, k = -2-√2

$\textbf { HOPE IT HELP YOU }$