Question

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

• A. $k=-2-\sqrt{2}$
• B. $k=-2+\sqrt{2}$
• C. $k=\sqrt{2}-1$
• D. $k=\sqrt{2}+1$

Answer 1

The correct option is C $k=\sqrt{2}-1$

Since, $x-1$ is factor of $p\left(x\right)=k{x}^2-\sqrt{2}x+1$

$\Rightarrow p\left(1\right)=0$

Now,$p\left(1\right)=k(1{)}^2-\sqrt{2}(1)+1=k-\sqrt{2}+1$

So, $k-\sqrt{2}+1=0$

$\Rightarrow k=\sqrt{2}-1$