The set of all values of k-1, for which the equation (3x^2+4x+3)^2-(k+1)(3x^2+4x+3)(3x^2+…

Question

The set of all values of $k>-1$, for which the equation $(3x^2+4x+3)^2-(k+1)(3x^2+4x+3)(3x^2+4x+2)+k(3x^2+4x+2)^2=0$ has real roots, is

Accepted Answer

Let $y=3x^2+4x+2$; the equation becomes $(y+1)^2-(k+1)y(y+1)+ky^2=0$, which simplifies to $y+1-ky=0 \implies y=\dfrac1{k-1}$.

So $3x^2+4x+2-\dfrac1{k-1}=0$ needs $D\ge0$: $16-24+\dfrac{12}{k-1}\ge0 \implies \dfrac3{k-1}\ge2 \implies \dfrac{2k-5}{k-1}\le0 \implies k\in\left(1,\dfrac52\right]$.

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