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प्रश्न
Solve the equation `9x^2 + (3x)/4 + 2 = 0`, if possible, for real values of x.
उत्तर
`9x^2 + (3x)/4 + 2 = 0`
`=> (36x^2 + 3x + 8)/4 = 0`
`=> 36x^2 + 3x + 8 = 0`
Here a = 36, b = 3 and c = 8
`∴ x = (-b +- sqrt(b^2 - 4ac))/(2a)`
`= (-3 +- sqrt((3)^2 - 4 xx 36 xx 8))/(2 xx 36)`
`= (-3 +- sqrt(9 - 1152))/72`
`= (-3 +- sqrt(-1143))/72`
Since `sqrt(-1143)` is not possible, we cannot solve the given equation for x.
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