Question

find nature of the roots 13√3x²+10x+√3=0

Answer 1

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Here's your answer...

The nature of the roots of a quadratic equation can be found by finding its discriminant.
D=b²-4ac
Here...
$a = 13 \sqrt{3} \\ b = 10 \\ c = \sqrt{3} \\ {b}^{2} - 4ac = {10}^{2} - 4 \times 13 \sqrt{3} \times \sqrt{3} \\ = 100 - 156 \\ = - 56$
Since -56<0, the discriminant is negative, and the roots of the equation will be imaginary.
(or complex at least)

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