# How to decide the number of solutions.. How to decide the number of solutions of equation 2x^2-square root3*x-6=0 ?

*print*Print*list*Cite

### 2 Answers

You have to remember that the number of solutions of an equation is equal to the highest power of the variable. A linear variable has 1 solution; a quadratic equation has 2 solutions, etc. The number of solutions may in some cases appear to be less than the highest power of the variable if some or all of them are equal.

For the equation you have given 2x^2- sqrt 3 * x - 6 = 0, the two solutions are:

[sqrt 3 + sqrt (3 + 48)]/4 = sqrt 3 / 4 + sqrt 51 / 4

and sqrt 3 / 4 - sqrt 51 / 4

The number of solutions could be equal to the maximum power of the variable x.

In this case, the maximum number of solutions cannot be greater than 2.

Now, we'll decide if we have 2 different solutions, 2 equal solutions or 2 complex solutions.

The character of solutions is given by the discriminant of the quadratic.

delta = b^2 - 4ac

a,b,c, are the coefficients of the given quadratic.

delta = 3 - 4*2*(-6)

delta = 3 + 48

delta = 51 > 0

Since delta is positive, the quadratic wil have 2 real different solutions.

So, the number of solutions is 2.