For what values of m are the roots of $x^2 +2x+3 = m(2x+1)$ real and positive

Question

I am only able to show that to be real, $m <-1$ or $m\geq2$ Don't know how to finish solution Answer is $2 \leq m < 3$

So far:

After expanding and factorising,

$x^2 + 2(1-m)x + (3-m) = 0 $

Roots are real if discriminant $\ge 0 $

i.e. $4(m-2)(m+1)>=0 $

Therefore $m ≤ -1 \text{ or } m \ge 2$ for roots to be real

However I don't know how to find m for roots to be positive

Answer 1

For the equation to have positive and real roots, two extra conditions are needed $$2m-2>0$$ $$3-m>0$$ $$\therefore 1<m<3$$ Taking the intersection of the intervals found, we will get $$\therefore 2 \leq m <3$$ :)

Answer 2

$x^2 +2x+3 = m(2x+1)$

$x^2 +2(1-m)x+3-m = 0$

$D=b^2-4ac$

$D=4(1-m)^2-4(3-m)$

$D=4(1+m^2-2m)-12+4m$

$D=4+4m^2-8m)-12+4m$

$D=4m^2-4m-8$

For $D \geq 0$,you will find real roots

roots are real for $m ≤ -1 \text{ or } m \ge 2$------(1)

$ x = (m-1) \pm \sqrt{m^2-m-2} $

roots are positive if $(m-1)>\sqrt{m^2-m-2}$

$m<3$---------------(2)

Combining (1) and (2) condition ,we get

roots are real and positive if $ 2 \leq m<3$

Answer 3

Notice your equation is the same as $$x^2 + x(2 - 2m) + (3 -m) = 0$$

Now, by the quadratic equation, the points where this parabola is zero, that means the zeros of the equations are:

$$ x = \frac{2m - 2 \pm \sqrt{4 - 8m + 4m^2 -12 +4m}}{2} $$

We only need to care about the determinant $\Delta(m) = 4m^2 -4m - 8$. You can see this equation geometrically or algebraically, whichever suits your tastes. But The determinant has the following properties: If the $\Delta$ of a quadration equation iz $0$, then you get unique solution. if $\Delta < 0 $, then we have no real solution. If $\Delta > 0$, then we have 2 solutions. Use this info to solve your problem.

HINT: Graph $\Delta(m) = 4m^2 -4m - 8$

Answer 4

Denoting the discriminant by $D(m) := 4(m+1)(m-2)$, we have for the roots $$ x_{1,2} = (m-1) \pm \sqrt{D(m)} $$ as you say correctly, the roots are real for $D(m) \ge 0$. And the roots are both positive if $0 \le \sqrt{D(m)} \le m-1$, so $m \ge 1$ and $D(m) \le (m-1)^2$ must hold. Now try to solve the last inequality for $m$.