Question

If ${x^2} + xy + xz = 135$, ${y^2} + yz + xy = 351$ and ${z^2} + xz + yz = 243$, then what is the value of ${x^2} + {y^2} + {z^2}$?
A. $300$
B. $275$
C. $250$
D. $225$

Answer 1

Hint: In this question, we are given three equations. Using these three equations, we have to find the value of ${x^2} + {y^2} + {z^2}$. Start by adding the three equations. This is done because after adding, it forms a formula - ${(x + y + z)^2}$. Then, we will find the value of the three valuables using the formula and then we will find the value of ${x^2} + {y^2} + {z^2}$ by squaring and adding the individual values.

Formula used: ${(x + y + z)^2} = {x^2} + {y^2} + {z^2} + 2(xy + xz + yz)$

Complete step-by-step answer:
We are given three equations. First, we will number the equations to avoid any confusions.
$ \Rightarrow {x^2} + xy + xz = 135$ ………..…. (1)
$ \Rightarrow {y^2} + xy + yz = 351$ …………….. (2)
$ \Rightarrow {z^2} + xz + yz = 243$ ……………... (3)
We will start by adding the three given equations (1) + (2) + (3),
$ \Rightarrow {x^2} + {y^2} + {z^2} + 2(xy + xz + yz)$=$135 + 351 + 243$ ……………….. (4)
On observation, we can notice that it is forming the formula of ${(x + y + z)^2}$.
$ \Rightarrow {(x + y + z)^2} = {x^2} + {y^2} + {z^2} + 2(xy + xz + yz)$ ………………….. (5)
On combining equation (4) and (5),
$ \Rightarrow {(x + y + z)^2} = 729 = {(27)^2}$
Square rooting both the sides,
$ \Rightarrow (x + y + z) = 27$
Taking $x$ common from equation (1),
$ \Rightarrow x(x + y + z) = 135$
Putting $(x + y + z) = 27$,
$ \Rightarrow x(27) = 135$
Finding the value of $x$,
$ \Rightarrow x = \dfrac{{135}}{{27}} = 5$
Taking $y$ common from equation (2),
$ \Rightarrow y(x + y + z) = 351$
Putting $(x + y + z) = 27$,
$ \Rightarrow y(27) = 351$
Finding the value of $y$,
$ \Rightarrow y = \dfrac{{351}}{{27}} = 13$
Taking $z$ common from equation (3),
$ \Rightarrow z(x + y + z) = 243$
Putting $(x + y + z) = 27$,
$ \Rightarrow z(27) = 243$
Finding the value of $z$,
$ \Rightarrow z = \dfrac{{243}}{{27}} = 9$
Now, we have the required values.
$ \Rightarrow x = 5,y = 13,z = 9$
Now, the value of ${x^2} + {y^2} + {z^2}$ can be founded by squaring and adding the individual terms.
$ \Rightarrow {x^2} + {y^2} + {z^2} = {5^2} + {13^2} + {9^2}$
Simplifying,
$ \Rightarrow {x^2} + {y^2} + {z^2} = 25 + 169 + 81$
$ \Rightarrow {x^2} + {y^2} + {z^2} = 275$

Therefore, our answer is (B) $275$.

Note: Square root must always be positive. Negative square root is not defined.
Some other basic formula:
${(x+y)^2}=x^2 + y^2 + 2xy$
${(x-y)^2}=x^2 + y^2 - 2xy$