Question

If \[\sqrt{x}+\dfrac{1}{\sqrt{x}}=\sqrt{6}\] , then the value of \[\left( {{x}^{4}}-190+\dfrac{1}{{{x}^{4}}} \right)\] is

Answer 1

Hint: To solve the question, we have to square the given equation until we get an expression in the terms of \[{{x}^{4}}\] . To solve further, apply the formula of square of sum of two variables and properties and formulae of algebraic exponents and powers.

Complete step by step solution:

The given expression is \[\sqrt{x}+\dfrac{1}{\sqrt{x}}=\sqrt{6}\]
On squaring on both sides of the equation, we get
$\Rightarrow$ \[{{\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)}^{2}}={{\left( \sqrt{6} \right)}^{2}}\]
$\Rightarrow$ \[{{\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)}^{2}}=6\]
We know the formula \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
Thus, by applying the formula to the above equation we get
$\Rightarrow$ \[{{\left( \sqrt{x} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{x}} \right)}^{2}}+2\sqrt{x}\times \dfrac{1}{\sqrt{x}}=6\]
$\Rightarrow$ \[{{\left( \sqrt{x} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{x}} \right)}^{2}}+2(1)=6\]
We know \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\]
Thus, by applying the formula to the above equation we get
$\Rightarrow$ \[x+\left( \dfrac{1}{x} \right)+2=6\]
$\Rightarrow$ \[x+\dfrac{1}{x}=6-2\]
$\Rightarrow$ \[x+\dfrac{1}{x}=4\]
On squaring on both sides of the equation, we get
$\Rightarrow$ \[{{\left( x+\dfrac{1}{x} \right)}^{2}}={{\left( 4 \right)}^{2}}\]
$\Rightarrow$ \[{{\left( x+\dfrac{1}{x} \right)}^{2}}=16\]
Thus, by applying the formula to the above equation we get
$\Rightarrow$ \[{{\left( x \right)}^{2}}+{{\left( \dfrac{1}{x} \right)}^{2}}+2x\times \dfrac{1}{x}=16\]
$\Rightarrow$ \[{{x}^{2}}+\left( \dfrac{1}{{{x}^{2}}} \right)+2(1)=16\]
$\Rightarrow$ \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}=16-2\]
$\Rightarrow$ \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}=14\]
On squaring on both sides of the equation, we get
$\Rightarrow$ \[{{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}={{\left( 14 \right)}^{2}}\]
$\Rightarrow$ \[{{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}=196\]
Thus, by applying the formula to the above equation we get
$\Rightarrow$ \[{{\left( {{x}^{2}} \right)}^{2}}+{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{2}}+2{{x}^{2}}\times \dfrac{1}{{{x}^{2}}}=196\]
$\Rightarrow$ \[{{\left( {{x}^{2}} \right)}^{2}}+{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{2}}+2(1)=196\]
We know \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\]
Thus, by applying the formula to the above equation we get

$\Rightarrow$ \[{{x}^{4}}+\left( \dfrac{1}{{{x}^{4}}} \right)+2=196\]
$\Rightarrow$ \[{{x}^{4}}+\dfrac{1}{{{x}^{4}}}=196-2\]
$\Rightarrow$ \[{{x}^{4}}+\dfrac{1}{{{x}^{4}}}=194\]
By rearranging the terms of the equation, we get
$\Rightarrow$ \[{{x}^{4}}+\dfrac{1}{{{x}^{4}}}=190+4\]
$\Rightarrow$ \[{{x}^{4}}+\dfrac{1}{{{x}^{4}}}-190=4\]
Thus, the value of \[\left( {{x}^{4}}-190+\dfrac{1}{{{x}^{4}}} \right)\] is 4.

Note: The possibility of mistake can be not squaring the given equation which is required to get an expression in the terms of \[{{x}^{4}}\] . The other possibility of mistake can be not applying the formula of square of sum of two variables and properties and formulae of algebraic exponents and powers. The other possibility of mistake can be the calculation mistake since the procedure of solving involves various calculations.