Determine all the values of the parameter m for which the operation: $4 x^{2} - 6 m x + (2 m + 3) (m - 3)$ $4x^2-6mx+(2m+3)(m-3)$ has TWO different real solutions. $x_{1} < x_{2}$ $x_1<x_2$ $(4 x_{1} - 4 x_{2} - 1) (4 x_{1} - 4 x_{2} + 1) < 0$ $(4x_1-4x_2-1)(4x_1-4x_2+1)<0$ ?

Question

$4 x^{2} - 6 m x + (2 m + 3) (m - 3)$ $4x^2-6mx+(2m+3)(m-3)$

$x_{1} < x_{2}$ $x_1<x_2$

$(4 x_{1} - 4 x_{2} - 1) (4 x_{1} - 4 x_{2} + 1) < 0$ $(4x_1-4x_2-1)(4x_1-4x_2+1)<0$

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Answer 1 · 2017-05-12T11:58:09

$m < - \frac{13}{2}$ $m < -13/2$ or $- \frac{1}{6} < m$ $-1/6 < m$

Solving for $x$ $x$

$4 x^{2} - 6 m x + (2 m + 3) (m - 3) = 0$ $4 x^2 - 6 m x + (2 m + 3) (m - 3) = 0$

we obtain

${(x_1=1/2(m-3)),(x_2=1/2(2m+3)):}$

so

$(4 x_1 - 4 x_2 - 1) (4 x_1 + 4 x_2 + 1) = -(12m^2+80m+13) < 0$

so we will determine $m$ such that

$12m^2+80m+13 > 0$

so

$m < -13/2$ or $-1/6 < m$