Find the value of $k$ , if $x - 1$ is a factor of $p (x)$ in each of the following cases:
$p (x) = 2 x^{2} + k x + \sqrt{2}$

k = - 2 - \sqrt{2}

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k = - 2 + \sqrt{2}

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k = 2 + \sqrt{2}

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k = 2 - \sqrt{2}

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Solution

The correct option is C $k = - 2 - \sqrt{2}$

Since, $x - 1$ is factor of $p (x) = 2 x^{2} + k x + \sqrt{2}$
$\Rightarrow p (1) = 0$
Now, $p (1) = 2 (1)^{2} + k (1) + \sqrt{2} = k + 2 + \sqrt{2}$
So, $k + 2 + \sqrt{2} = 0$
$\Rightarrow k = - 2 - \sqrt{2}$

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Similar questions

Q. Find the value of k, if x-1 is a factor of p(x) in each of the following cases

p (x) = {2 x}^{2} + k x + \sqrt{2}

Find the value of k, if (x −1) is a factor of p(x) in the following case:
$p (x) = {2 x}^{2} + k x + \sqrt{2}$

Q. Find the value of

k

, if

x - 1

is a factor of

p (x)

in each of the following cases:
(i)

p (x) = x^{2} + x + k

(ii)

p (x) = 2 x^{2} + k x + \sqrt{2}

(iii)

p (x) = k x^{2} - \sqrt{2} x + 1

(iv)

p (x) = k x^{2} - 3 x + k

Find the value of k, if (x −1) is a factor of $p (x) = {2 x}^{2} + k x + \sqrt{2}$

Find the value of k, if (x −1) is a factor of p(x) in each of the following cases:
$(i) p (x) = x^{2} + x + k$
$(i i) p (x) = {2 x}^{2} + k x + \sqrt{2}$ [4 MARKS]

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Find the value of k, if x−1 is a factor of p(x) in each of the following cases:p(x)=2x2+kx+√2

The correct option is C k=−2−√2Since, x−1 is factor of p(x)=2x2+kx+√2⇒p(1)=0Now, p(1)=2(1)2+k(1)+√2=k+2+√2So, k+2+√2=0⇒k=−2−√2

Find the value of $k$ , if $x - 1$ is a factor of $p (x)$ in each of the following cases:
$p (x) = 2 x^{2} + k x + \sqrt{2}$

The correct option is C $k = - 2 - \sqrt{2}$

Since, $x - 1$ is factor of $p (x) = 2 x^{2} + k x + \sqrt{2}$
$\Rightarrow p (1) = 0$
Now, $p (1) = 2 (1)^{2} + k (1) + \sqrt{2} = k + 2 + \sqrt{2}$
So, $k + 2 + \sqrt{2} = 0$
$\Rightarrow k = - 2 - \sqrt{2}$