The discriminant shall be $$16(abcd)^2-4a^4c^4-4a^4d^4-4b^4c^4-4b^4d^4$$ $$-4(b^2d^2+a^2c^2)^2-4(a^2d^2+b^2d^2)^2$$ which is clearly a negative value. So the roots should be imaginary. But the answer given is real and equal. How is that possible?
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It will be $$-4(a^2b^2-c^2d^2)^2-4(a^2d^2-b^2c^2)^2\le0$$
lab bhattacharjee
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Why? Since the square will will always be positive and multiplied by a negative 4, the value will be negative – Aditya Nov 17 '19 at 07:40
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@Aditya, Sorry for the typo – lab bhattacharjee Nov 17 '19 at 07:43
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Okay, so the roots will still be imaginary right? – Aditya Nov 17 '19 at 07:50
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1@Aditya, unless both perfect squares are $0$ – lab bhattacharjee Nov 17 '19 at 08:02
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Your discriminant is written slightly wrong it should be $$-4(b^2d^2 - a^2c^2)^2 -4(a^2d^2 - b^2c^2)^2.$$ Further as per the equation you have written you are surely correct that it will have imaginary roots. May be the answer given is wrong or some more information may be provided. Hope this helps.
Aditya
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Sameer Nilkhan
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The discriminant is given by $$-a^4c^4-a^4d^4+4a^2b^2c^2d^2-b^4c^4-b^4a^4$$ and now completing the squares.
Dr. Sonnhard Graubner
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