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Question
In the following, determine whether the given quadratic equation have real roots and if so, find the roots:
`2x^2+5sqrt3x+6=0`
Solution
We have been given, `2x^2+5sqrt3x+6=0`
Now we also know that for an equation ax2 + bx + c = 0, the discriminant is given by the following equation:
D = b2 - 4ac
Now, according to the equation given to us, we have,a = 2, `b=5sqrt3` and c = 6.
Therefore, the discriminant is given as,
`D=(5sqrt3)^2-4(2)(6)`
= 75 - 48
= 27
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real roots.
Now, the roots of an equation is given by the following equation,
`x=(-b+-sqrtD)/(2a)`
Therefore, the roots of the equation are given as follows,
`x=(-(5sqrt3)+-sqrt27)/(2(2))`
`=(-5sqrt3+-3sqrt3)/4`
Now we solve both cases for the two values of x. So, we have,
`x=(-5sqrt3+3sqrt3)/4`
`=(-sqrt3)/2`
Also,
`x=(-5sqrt3-3sqrt3)/4`
`=-2sqrt3`
Therefore, the roots of the equation are `(-sqrt3)/2` and `-2sqrt3`.
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