# How do i find the "EXACT" no of real roots of a equation?No HIT AND TRIAL PLEASE(x^4-4x^3+3x^2+2x+6=0)I solved it using descarte's rule of sings .found 1 negative root and 3 positive roots or 1...

How do i find the "EXACT" no of real roots of a equation?No HIT AND TRIAL PLEASE(x^4-4x^3+3x^2+2x+6=0)

I solved it using descarte's rule of sings .found 1 negative root and 3 positive roots or 1 negative,1 positive and 2 complex)How do i make sure for no of positive roots without hit and trial or solving ...Any shortcuts......

### 1 Answer | Add Yours

You need to evaluate the derivative of the equation of function such that:

`4x^3 - 12x^2 + 6x+ 2 = 0`

You need to divide all equation by 2 such that:

`2x^3 - 6x^2 + 3x+ 1 = 0`

You need to look the zeroes of derivative among the following fractions: `+-1/1 ; +-1/2` .

`x = -1 =gt -2 - 6 - 3 + 1 != 0`

`x = 1 =gt 2 - 6 + 3 + 1 = 0`

Notice that x = 1 cancels the equation, hence you may write the equation of derivative such that:

`2x^3 - 6x^2 + 3x + 1 = (x - 1)(ax^2 + bx+ c)`

`2x^3 - 6x^2 + 3x + 1 = ax^3 + bx^2 + cx - ax^2 - bx - c`

Equating the coefficients of like powers yields:

`a = 2`

`b -a = -6 =gt b - 2 = -6 =gt b = -4`

`c - b = 3 =gt c + 4 = 3 =gt c = -1`

`2x^3 - 6x^2 + 3x + 1 = (x - 1)(2x^2 - 4x - 1)`

You need to find the zeroes of `2x^2 - 4x - 1 = 0` such that:

`x_(1,2) = (4 +- sqrt(16 + 8))/4`

`x_(1,2) = (4 +-2 sqrt6)/4`

`x_(1,2) = (2 +- sqrt6)/2`

You need to evaluate the value of function for each root of derivative such that:

`x =1 =gt1-4+3+2+6 = 8 ` ( + sign)

`x =-1 =gt 1+4+3-2+6 =12 gt 0`

`x = 2 =gt 16-32+12+4+6 gt 0`

**Hence, over the real set, the values of the function are positive, hence, it means that the graph is floating above x axis and the equation has no real roots.**

**Sources:**