Find the nature of the roots of the quadratic equation
13root3x^2+10x+root3=0
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here the answer is...
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Here a= 13√3
b= 10
c = √3
so, b²-4ac = (10)² - 4 × 13√3 × √3
= 100 - 156
= -56
Since b²-4ac < 0 , the quadratic equation has no real roots.
b= 10
c = √3
so, b²-4ac = (10)² - 4 × 13√3 × √3
= 100 - 156
= -56
Since b²-4ac < 0 , the quadratic equation has no real roots.
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