First slide

Theory of equations

Question

The number of pairs (x, y) which will satisfy the equation $x^{2} - xy + y^{2} = 4 (x + y - 4)$ is

Moderate

A

1
B

2
C

4
D

None of these

Solution

we have, $x^{2} - xy + y^{2} = 4 (x + y - 4)$
$\begin{aligned} \Rightarrow x^{2} - x (y + 4) + y^{2} - 4 y + 16 & = 0 \\ ∵ x & \in R \\ ∵ (- (y + 4))^{2} - 4 \cdot 1 \cdot (y^{2} - 4 y + 16) & \geq 0 \end{aligned}$
$\begin{array}{lr} \Rightarrow & y^{2} + 8 y + 16 - 4 y^{2} + 16 y - 64 \geq 0 \\ \Rightarrow & 3 y^{2} - 24 y + 48 \leq 0 \\ \Rightarrow & y^{2} - 8 y + 16 \leq 0 \Rightarrow (y - 4)^{2} \leq 0 \\ ∴ & (y - 4)^{2} = 0 \end{array}$
$\begin{aligned} ∴ y & = 4 \\ x^{2} - 4 x + 16 & = 4 (x + 4 - 4) \\ x^{2} - 8 x + 16 & = 0 \\ (x - 4)^{2} & = 0 \\ x & = 4 \end{aligned}$
Number of pairs is 1 i.e., (4, 4).

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