Question

Solve the following equations:
\[{{x}^{4}}+{{y}^{4}}=272\]
\[x-y=2\]
(This question has multiple correct options)
(a) (4, 2)
(b) (3, 4)
(c) (– 2, – 4)
\[\left( \text{d} \right)\text{ }\left( \dfrac{\sqrt{3}+2}{4},\dfrac{1}{2} \right)\]

Answer 1

Hint: At first, consider the equation \[{{x}^{4}}+{{y}^{4}}=272\] and use the identities \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] and \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] to convert in terms of xy and hence find its value. Lastly, use \[{{\left( x+y \right)}^{2}}={{\left( x-y \right)}^{2}}+4xy\] to find the value of x and y.

Complete step-by-step answer:
In this question, we are given two equations containing the variables x and y only. The equations are \[{{x}^{4}}+{{y}^{4}}=272\] and x – y = 2 and from these, we have to find the values of x and y. So, at first, we take the equation \[{{x}^{4}}+{{y}^{4}}=272.\]
Now, we will add and subtract by \[2{{x}^{2}}{{y}^{2}}.\] We get,
\[{{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}-2{{x}^{2}}{{y}^{2}}=272\]
The result remains unchanged.
Now, we will apply the identity,
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
Now, apply \[{{a}^{2}}+{{b}^{2}}+2ab\text{ as }{{\left( a+b \right)}^{2}}\]
So, we can write the equation as,
\[{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-2{{x}^{2}}{{y}^{2}}=272\]
We can write, \[{{x}^{2}}+{{y}^{2}}\] as \[{{x}^{2}}+{{y}^{2}}-2xy+2xy.\] So, we get,
\[{{\left\{ \left( {{x}^{2}}+{{y}^{2}}-2xy \right)+2xy \right\}}^{2}}-2{{x}^{2}}{{y}^{2}}=272\]
Now, we can use the identity \[\left( {{x}^{2}}+{{y}^{2}}-2xy \right)={{\left( x-y \right)}^{2}}\] and write the equation as,
\[{{\left\{ {{\left( x-y \right)}^{2}}+2xy \right\}}^{2}}-2{{x}^{2}}{{y}^{2}}=272\]
We know that the value of x – y =2. So, we will apply it and get,
\[{{\left\{ {{2}^{2}}+2xy \right\}}^{2}}-2{{x}^{2}}{{y}^{2}}=272\]
\[\Rightarrow {{\left( 4+2xy \right)}^{2}}-2{{x}^{2}}{{y}^{2}}=272\]
Now, we will apply the identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] where a is 4 and b is 2xy. So, we get,
\[16+16xy+4{{x}^{2}}{{y}^{2}}-2x{{y}^{2}}=272\]
\[\Rightarrow 2{{x}^{2}}{{y}^{2}}+16xy-256=0\]
\[\Rightarrow {{x}^{2}}{{y}^{2}}+8xy-128=0\]
We can write it as,
\[{{x}^{2}}{{y}^{2}}-8xy+16xy-128=0\]
\[\Rightarrow xy\left( xy-8 \right)+16\left( xy-8 \right)=0\]
After factorizing, we get,
\[\left( xy+16 \right)\left( xy-8 \right)=0\]
So, the values of xy are – 16 or 8.
Now, as we know that the value of (x – y) as 2 and xy as either – 16 or 8. So, we will try to find the values of x + y using the two cases of (xy) values.
Here, we will use the formula,
\[{{\left( x+y \right)}^{2}}={{\left( x-y \right)}^{2}}+4xy\]
For the value, x – y = 2 and xy = – 16, we get,
\[{{\left( x+y \right)}^{2}}={{2}^{2}}+4\times \left( -16 \right)\]
\[\Rightarrow {{\left( x+y \right)}^{2}}=4-64=-60\]
which is not possible, hence this case is invalid.
Now, for the value x – y = 2 and xy = 8, we get,
\[{{\left( x+y \right)}^{2}}={{2}^{2}}+4\times \left( 8 \right)\]
\[\Rightarrow {{\left( x+y \right)}^{2}}=4+32=36\]
So, the values of (x + y) are 6 and – 6.
Now, we will find the values of x and y for the cases of value (x + y) which are 6 and – 6. Therefore, x + y = – 6 and x – y = 2. We will add both the equations, we get,
\[\left( x+y \right)+\left( x-y \right)=-6+2\]
\[\Rightarrow 2x=-4\]
\[\Rightarrow x=-2\]
We know that, x + y = – 6 and x = – 2. So, we can say,
– 2 + y = – 6
y = – 4
Now, for x + y = 6 and x – y = 2.
Here, x + y = 6, x – y = 2. We will add both the equations, we get,
\[\left( x+y \right)+\left( x-y \right)=6+2\]
\[\Rightarrow 2x=8\]
\[\Rightarrow x=4\]
We know that x + y = 6 and x = 4. So, we can say,
4 + y = 6
y = 2
Hence, the values of (x, y) are (4, 2) and (– 2, – 4).
So, the correct options are (a) and (c).

Note: We can also solve this problem by just referring to the options. Instead of solving the problem, just substitute the values of x and y in the equation and check which one satisfies. And the one which satisfies wil be the required result. There is a possibility of having more than one option as the correct answer therefore one must check all the options.