Question

If x – y + z = 5 and \[{{x}^{2}}{{y}^{2}}+{{z}^{2}}~=49\] , then the value of \[zx\text{ }\text{ }xy\text{ }\text{ }yz~\] is
a). 12
b). -12
c). 37
d). -37

Answer 1

Hint: To solve the question, we have to apply the formula for the square of sum of two variables. By solving the obtained expression by using the above formula and by substituting the given values of expression, we can arrive at the answer.

Complete step-by-step answer:
Consider the given equation x – y + z = 5
By squaring on both sides of equation we get
\[\begin{align}
  & {{\left( x-y+z \right)}^{2}}={{\left( 5 \right)}^{2}}~ \\
 & {{\left( x-y+z \right)}^{2}}=25~ \\
\end{align}\]
Let k be equal to x-y. By substituting this value in the above equation, we get
\[~{{\left( k+z \right)}^{2}}=25\]
We know the formula \[~{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
On comparison with our equation, we get
a = k, b = z
By substituting the above values in the formula, we get
\[~{{k}^{2}}+{{z}^{2}}+2kz=25\]
By substituting the value of k in the above equation, we get
\[\begin{align}
  & {{(x-y)}^{2}}+{{z}^{2}}+2(x-y)z=25 \\
 & {{(x-y)}^{2}}+{{z}^{2}}+2(xz-yz)=25 \\
\end{align}\]
By applying the formula \[~{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] to the above equation, we get
\[~{{x}^{2}}+{{y}^{2}}-2xy+{{z}^{2}}+2(xz-yz)=25\]
\[~{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2(xz-yz-xy)=25\] …… (1)
The given value of \[{{x}^{2}}{{y}^{2}}+{{z}^{2}}~\] is equal to 49.
By substituting the above value in the equation (1), we get
The given value of \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}~\] is equal to 49.
\[\begin{align}
  & ~49+2(xz-yz-xy)=25 \\
 & 2(xz-yz-xy)=25-49 \\
 & 2(xz-yz-xy)=-24 \\
 & xz-yz-xy=\dfrac{-24}{2} \\
 & xz-yz-xy=-12 \\
\end{align}\]
Thus, the value of \[zx\text{ }\text{ }xy\text{ }\text{ }yz~\] is equal to -12
Hence, option (b) is the right choice.
Note: The possibility of mistake is not applying the appropriate formula to solve the equations. The other possibility of mistake can be the calculation mistake since the procedure of solving involves more substitution and calculation. The alternative way of solving can be applying the direct formula \[{{\left( x+y+z \right)}^{2}}=~{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2(xz+yz+xy)\] . Then, by changing the sign of y and substituting the given values we can calculate the required answer.