√3x^2+√3x+6√3. Find the roots of this equation
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Answered by
1
D = b ^2 - 4ac
= ( √3 )^2 - 4 * √3 * 6 √3
= 3 - 18 * 4
= - 72
Since , the value of D is. negative , we can't find the real roots..
= ( √3 )^2 - 4 * √3 * 6 √3
= 3 - 18 * 4
= - 72
Since , the value of D is. negative , we can't find the real roots..
agw18112002:
But the furthur you solve it comes like 3-18×4
3 - 18* 4 = 3 - 72
it's negative
Answered by
0
X=-b+/-√Dwhole upon 2a it is quadratic formula or shridhar acharya formula
x=-√3+/-√3-72whole upon 2√3
so the roots are imaginary okk
x=-√3+/-√3-72whole upon 2√3
so the roots are imaginary okk
Dear it has imaginary roots
okk
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