First slide

Theory of equations

Question

$The roots of the equation (a^{4} + b^{4}) x^{2} + 4 a b c d x + (c^{4} + d^{4}) = 0$

Moderate

A

cannot be different, if real
B

are always real
C

are always imaginary
D

None of these

Solution

The discriminant of the given equation is
$\begin{array}{l} D = 16 a^{2} b^{2} c^{2} d^{2} - 4 (a^{4} + b^{4}) (c^{4} + d^{4}) \\ = - 4 [(a^{4} + b^{4}) (c^{4} + d^{4}) - 4 a^{2} b^{2} c^{2} d^{2}] \\ = - 4 [a^{4} c^{4} + a^{4} d^{4} + b^{4} c^{4} + b^{4} d^{4} - 4 a^{2} b^{2} c^{2} d^{2}] \\ = - 4 [(a^{4} c^{4} + b^{4} d^{4} - 2 a^{2} b^{2} c^{2} d^{2}) + (a^{4} d^{4} + b^{4} c^{4} - 2 a^{2} b^{2} c^{2} d^{2})] \\ = - 4 [{(a^{2} c^{2} - b^{2} d^{2})}^{2} + {(a^{2} d^{2} - b^{2} c^{2})}^{2}] - - - - - (1) \end{array}$
$If roots of the given equation are real, D \geq 0$
$\begin{array}{l} \Rightarrow - 4 [{(a^{2} c^{2} - b^{2} d^{2})}^{2} + {(a^{2} d^{2} - b^{2} c^{2})}^{2}] \geq 0 \\ \Rightarrow {(a^{2} c^{2} - b^{2} d^{2})}^{2} + {(a^{2} d^{2} - b^{2} c^{2})}^{2} \leq 0 \\ \Rightarrow {(a^{2} c^{2} - b^{2} d^{2})}^{2} + {(a^{2} d^{2} - b^{2} c^{2})}^{2} = 0 - - - - - - - - (2) \end{array}$
$[∵ Sum of two positive quantities cannot be negative]$
$From (1) and (2), we get D = 0$
Hence, the roots of the given quadratic equation are not
different if they are real.

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