First slide

Theory of equations

Question

$Number of ordered pairs (x, y) of real numbers satisfying the equation 2 (x^{4} - 2 x^{2} + 3) (y^{4} - 3 y^{2} + 4) = 7, is$

Moderate

A

2
B

4
C

6
D

8

Solution

$\begin{array}{l} 2 (x^{4} - 2 x^{2} + 3) (y^{4} - 3 y^{2} + 4) = 7 \\ \Rightarrow 2 ({(x^{2} - 1)}^{2} + 2) ({(y^{2} - \frac{3}{2})}^{2} + \frac{7}{4}) = 7 \end{array}$
Now, least value of L.H.S. is 7.
$\begin{array}{l} Hence, equality holds when x^{2} = 1 and y^{2} = \frac{3}{2} . \\ \Rightarrow x = \pm 1 and y = \pm \frac{\sqrt{3}}{2} \end{array}$
Hence, 4 ordered pairs satisfy the equation.
$These are (1, \frac{\sqrt{3}}{2}), (1, \frac{- \sqrt{3}}{2}), (- 1, \frac{\sqrt{3}}{2}), (- 1, \frac{- \sqrt{3}}{2})$

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