Question

 In the following equation, find the value of $k$ for which the given value is a solution of the given equation, the value of $k$ is:
$k{{x}^{2}}+\sqrt{2x}-4=0$ , $x=\sqrt{2}$

Answer 1

Hint: For solving this question as it is given that $x=\sqrt{2}$ is a solution of the equation $k{{x}^{2}}+\sqrt{2x}-4=0$ so, we will directly substitute the value of $x=\sqrt{2}$ in the given equation and solve for the correct value of $k$ easily.

Complete step by step answer:
Given:
It is given that $x=\sqrt{2}$ is a solution of the equation $k{{x}^{2}}+\sqrt{2x}-4=0$ and we have to find the suitable value of $k$ .
Now, we can interpret the given data as for the value $x=\sqrt{2}$ , the value of the function $f\left( x \right)=k{{x}^{2}}+\sqrt{2x}-4$ is equal to zero. Which means $f\left( \sqrt{2} \right)=0$ . Then,
\[\begin{align}
  & f\left( x \right)=k{{x}^{2}}+\sqrt{2x}-4 \\
 & \Rightarrow f\left( \sqrt{2} \right)=k{{\left( \sqrt{2} \right)}^{2}}+\sqrt{2\times \sqrt{2}}-4=0 \\
 & \Rightarrow 2k+\sqrt{2\times {{2}^{0.5}}}-4=0 \\
 & \Rightarrow 2k+{{\left( {{2}^{1.5}} \right)}^{0.5}}-4=0 \\
\end{align}\]
Now, before we proceed we should know that if ${{a}^{b}}=c$ then, ${{c}^{d}}={{\left( {{a}^{b}} \right)}^{d}}={{a}^{bd}}$ . Then, in the above equation, we can write ${{\left( {{2}^{1.5}} \right)}^{0.5}}={{2}^{1.5\times 0.5}}$ . Then,
\[\begin{align}
  & 2k+{{\left( {{2}^{1.5}} \right)}^{0.5}}-4=0 \\
 & \Rightarrow 2k+{{2}^{1.5\times 0.5}}-4=0 \\
 & \Rightarrow 2k=4-{{2}^{0.75}} \\
 & \Rightarrow k=\dfrac{4-{{2}^{0.75}}}{2} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow k=2-\dfrac{1}{{{2}^{1-0.75}}} \\
 & \Rightarrow k=2-\dfrac{1}{{{2}^{0.25}}} \\
\end{align}\]
Now, we will take the value of $\dfrac{1}{{{2}^{0.25}}}$ to be 0.8409 approximately to calculate the approximate value of the unknown term $k$ . Then,
\[\begin{align}
  & k=2-\dfrac{1}{{{2}^{0.25}}} \\
 & \Rightarrow k\approx 2-0.8409 \\
 & \Rightarrow k\approx 1.1591 \\
\end{align}\]
Now, from the above result, we conclude that if $x=\sqrt{2}$ is a solution of the equation $k{{x}^{2}}+\sqrt{2x}-4=0$ then the value of $k$ will be equal to 1.1591 approximately.
Thus, the required value of \[k=2-\dfrac{1}{{{2}^{0.25}}}\approx 1.1591\] .

Note: Here, the student should first understand and then proceed in the right direction to get the correct answer quickly. Moreover, though the question is very easy, but we should avoid making calculation mistakes while solving the question like when we put $x=\sqrt{2}$ , then one mistake is very common that we might take $\sqrt{2x}=\sqrt{2}\times \sqrt{2}$ . So, don’t make such mistakes while solving.