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If \[x+y=5\] and \[xy=4\], find \[{{x}^{3}}+{{y}^{3}}\].

seo-qna
Last updated date: 13th Jun 2024
Total views: 371.1k
Views today: 8.71k
Answer
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Hint: In solving the solution of this question, we will first substitute from both the equations. We will write the equation \[x+y=5\] by putting the term y in the left side of equation and the in the right side of the equation, we will put the term y and the constant term 5. From here, we will take the value of y and put this value of y in the equation \[xy=4\] and find the value of x. And, from there we will find the value of y by putting the value of x in any of the equations. After that, we will put the value of x and y in the term \[{{x}^{3}}+{{y}^{3}}\], then we will get the answer.

Complete step by step solution:
Let us solve this question.
In this question, we have given two equations that are \[x+y=5\] and \[xy=4\].
Here, we will first solve the equation \[x+y=5\].
We will take the value of x to the right side of equation \[x+y=5\]
So, the equation \[x+y=5\] can also be written as
\[\Rightarrow y=5-x\]
Now, we will put this value of y (that is \[y=5-x\]) in the equation \[xy=4\].
After putting the value of y in the equation \[xy=4\], then we will get
\[\Rightarrow x\left( 5-x \right)=4\]
We can write the above equation as
\[\Rightarrow 5x-{{x}^{2}}=4\]
We can write the above equation as
\[\Rightarrow {{x}^{2}}-5x+4=0\]
As we know that
If the equation is \[a{{x}^{2}}+bx+c=0\], then the value of x will be
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Hence, the value of x in the equation \[{{x}^{2}}-5x+4=0\] will be
\[x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 1\times 4}}{2\times 1}\]
The above can also be written as
\[x=\dfrac{5\pm \sqrt{25-16}}{2}=\dfrac{5\pm \sqrt{9}}{2}=\dfrac{5\pm 3}{2}\]
Hence, the value of x will be \[\dfrac{5+3}{2}\] and \[\dfrac{5-3}{2}\]
Or, we can say that the values of x are 4 and 1.
If we will put the value of x as 4 in the equation \[xy=4\], then the value of y will be 1.
If we will put the value of x as 1 in the equation \[xy=4\], then the value of y will be 4.
For the value of x and y as 4 and 1 respectively, then the value of the term \[{{x}^{3}}+{{y}^{3}}\] will be \[{{4}^{3}}+{{1}^{3}}=65\]
For the value of x and y as 1 and 4 respectively, then the value of the term \[{{x}^{3}}+{{y}^{3}}\] will be \[{{1}^{3}}+{{4}^{3}}=65\]
Hence, we get that the value of the term \[{{x}^{3}}+{{y}^{3}}\] is 65.

Note:
We should know how to substitute a variable from the two equations, so that we can solve this type of question easily. We should know that if we have a given a quadratic equation in the form of \[a{{x}^{2}}+bx+c=0\], then the value of x for this type of equation will be \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
If we want to cross check if our answer is correct, then we can check the equations \[x+y=5\] and \[xy=4\] are satisfying or not by putting the value of x and y,
If we put the value of x as 4 or 1 and y as 1 or 4 in the equation \[x+y=5\] , then the equation will be like: 5=5, hence it satisfies.
And, if we put the value of x as 4 or 1 and y as 1 or 4 in the equation \[xy=4\], then the equation will be like: 4=4, hence it satisfies both the equation and says that answers are correct.