Question

If an algebraic expression is given as x \[=3-2\sqrt{2}\] , then find \[{{x}^{4}}+\dfrac{1}{{{x}^{4}}}\] ?

Answer 1

HINT: For evaluating this question, we have to know the following identity and that is \[\left( {{a}^{2}}+{{b}^{2}}+2ab \right)={{\left( a+b \right)}^{2}}\] Now, on observing the question, we can see that the expression that needs to be evaluated is having terms that are reciprocal of each other. So, we can see that if we multiply the two terms of the expression, then the thing that we get is 1.

Complete step-by-step solution -
As mentioned in the question, we have to find the value of the expression that is given in the question.
Now, as mentioned in the hint, we can try to write the expression in a form by which we can convert the given expression into the identity that is mentioned in the hint and we can do that as follows
$ {{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}={{x}^{4}}+\dfrac{1}{{{x}^{4}}}+2{{x}^{2}}\cdot \dfrac{1}{{{x}^{2}}} \\ $
 $\Rightarrow {{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}={{x}^{4}}+\dfrac{1}{{{x}^{4}}}+2 \\ $
 $\Rightarrow {{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}-2={{x}^{4}}+\dfrac{1}{{{x}^{4}}}\ \ \ \ \ ……...(a) \\ $
Now, again we can convert the obtained expression as follows
$ {{\left( x+\dfrac{1}{x} \right)}^{2}}={{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2{{x}}\cdot \dfrac{1}{{{x}}} \\ $
 $\Rightarrow {{\left( x+\dfrac{1}{x} \right)}^{2}}={{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2 \\ $
$\Rightarrow {{\left( x+\dfrac{1}{x} \right)}^{2}}-2={{x}^{2}}+\dfrac{1}{{{x}^{2}}} \\ $
Now, using the equation (a), we can write the above expression as follows
 $\Rightarrow {{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}-2={{x}^{4}}+\dfrac{1}{{{x}^{4}}}\ \ \ \ \ \ \ ………..(a) \\ $
 $\Rightarrow {{\left( x+\dfrac{1}{x} \right)}^{2}}-2={{x}^{2}}+\dfrac{1}{{{x}^{2}}} \\ $
 $\Rightarrow{{\left( {{\left( x+\dfrac{1}{x} \right)}^{2}}-2 \right)}^{2}}-2={{x}^{4}}+\dfrac{1}{{{x}^{4}}} \\ $
Now, we can put the value of x and hence, we will obtain the following expression
$\Rightarrow {{\left( {{\left( x+\dfrac{1}{x} \right)}^{2}}-2 \right)}^{2}}-2={{x}^{4}}+\dfrac{1}{{{x}^{4}}} \\ $
 $\Rightarrow {{\left( {{\left( 3-2\sqrt{2}+\dfrac{1}{3-2\sqrt{2}} \right)}^{2}}-2 \right)}^{2}}-2={{x}^{4}}+\dfrac{1}{{{x}^{4}}} \\ $
$\Rightarrow {{x}^{4}}+\dfrac{1}{{{x}^{4}}}={{\left( {{\left( 3-2\sqrt{2}+\dfrac{1}{3-2\sqrt{2}} \right)}^{2}}-2 \right)}^{2}}-2 \\ $
$\Rightarrow {{x}^{4}}+\dfrac{1}{{{x}^{4}}}={{\left( {{\left( \dfrac{{{\left( 3-2\sqrt{2} \right)}^{2}+1}}{3-2\sqrt{2}} \right)}^{2}}-2 \right)}^{2}}-2 \\ $
$\Rightarrow {{x}^{4}}+\dfrac{1}{{{x}^{4}}}={{\left( {{\left( \dfrac{18-12\sqrt{2}}{3-2\sqrt{2}} \right)}^{2}}-2 \right)}^{2}}-2 \\ $
 $\Rightarrow {{x}^{4}}+\dfrac{1}{{{x}^{4}}}={{\left( {{\left( \dfrac{6\left( 3-2\sqrt{2} \right)}{3-2\sqrt{2}} \right)}^{2}}-2 \right)}^{2}}-2 \\ $
$\Rightarrow {{x}^{4}}+\dfrac{1}{{{x}^{4}}}={{\left( {{\left( 6 \right)}^{2}}-2 \right)}^{2}}-2 \\ $
 $\Rightarrow {{x}^{4}}+\dfrac{1}{{{x}^{4}}}={{\left( 36-2 \right)}^{2}}-2 \\ $
$\Rightarrow {{x}^{4}}+\dfrac{1}{{{x}^{4}}}={{\left( 34 \right)}^{2}}-2=1156-2=1154 \\ $

Hence, the value of the given expression is 1154.

NOTE: - The students can make an error if they don’t know about the property that is mentioned in the hint. For solving this question, one has to be able to identify the given expression and the identity into which it can be converted.