# For what values of m are the roots of x^2 +2x +3 =m(2x+1) real and positive. I rearranged the equation to x^2 -(2m-2)x+(m+3)=0 is that right??

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You need to consider the condition that requests that the roots `x_1` and `x_2` to be real, hence, you should consider the following condition regarding the radicand, such that:

`m^2 - m - 2 >= 0`

You need to attach the quadratic equation, such that:

`m^2 - m - 2= 0`

`m_(1,2) = (1+-sqrt(1+8))/2 => m_(1,2) = (1+-3)/2`

`m_1 = 2, m_2 = -1`

You need to notice that the quadratic expression is positive over the intervals `(-oo,-1]U[2,+oo)` .

You should consider the next condition regarding the roots, such that, `x_1` and `x_2` are both positive real roots:

`x_1*x_2 > 0 <=> {(x_1>0, x_2>0):}`

Hence, performing the product of the roots yields:

`(m - 1 + sqrt(m^2 - m - 2))(m - 1 - sqrt(m^2 - m - 2)) = (m - 1)^2 - (m^2 - m - 2)`

Replacing `(m - 1)^2 - (m^2 - m - 2)` for `x_1*x_2` in inequality `x_1*x_2 > 0` yields:

`(m - 1)^2 - (m^2 - m - 2) > 0`

You need to square such that:

`m^2 - 2m + 1 - m^2 + m + 2 > 0`

Reducing suplicate members yields:

`-m + 3 > 0 => -m > -3 => m < 3`

You need to consider the system of both conditions regarding the roots, such that:

`{(m in (-oo,-1]U[2,+oo)),(m in (-oo,3)):}`

The common interval that satisfies the request of the problem is found by performing the intersection of intervals such that:

`m in (-oo,-1]U[2,+oo) nn (-oo,3) => m in [2,3) => 2 <= m < 3`

**Hence, evaluating the interval of values of m for the roots `x_1, x_2` are both real positive, yields `2 <= m < 3` .**

You have the right idea in the first step, but you made a small calculation error:

`f(x)=x^2+2x-2xm+3-m=0`

`=x^2+(2-2m)x+(3-m)=0`

We can use the quadratic formula in order to find a function for x in terms of m:

a=1; b=2-2m; c=3-m

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`=(2m-2+-sqrt((2-2m)^2-4(1)(3-m)))/(2(1))`

`=(2m-2+-sqrt(4-8m+4m^2-12+4m))/2`

`=(2m-2+-sqrt(4(m^2-m-2)))/2`

`=m-1+-sqrt(m^2-m-2)gt0`

`m-1-sqrt(m^2-m-2)gt0`

`m-1gtsqrt(m^2-m-2)`

`m^2-2m+2gtm^2-m-2`

`-2m+mgt-2-2`

`-mgt-4`

`mgt4`

`m-1+sqrt(m^2-m-2)gt0`

`m-1gt-sqrt(m^2-m-2)`

This will result in the same solution.

Therefore, in order for the roots of the function to be both positive and real, m>4.