The set of all values of k -1,for which the | vedclass

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(A) Let $a = 3x^2 + 4x + 3$ and $b = 3x^2 + 4x + 2$. Note that $b = a - 1$.The given equation becomes $a^2 - (k + 1)ab + kb^2 = 0$.Factoring the quadr

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$(1, \frac{5}{2}]$

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(A) Let $a = 3x^2 + 4x + 3$ and $b = 3x^2 + 4x + 2$. Note that $b = a - 1$.The given equation becomes $a^2 - (k + 1)ab + kb^2 = 0$.Factoring the quadratic in terms of $a$ and $b$:$a^2 - kab - ab + kb^2 = 0$$a(a - kb) - b(a - kb) = 0$$(a - b)(a - kb) = 0$This gives two cases:$1) \; a = b$ $\Rightarrow 3x^2 + 4x + 3 = 3x^2 + 4x + 2$ $\Rightarrow 3 = 2$,which is impossible.$2) \; a = kb \Rightarrow 3x^2 + 4x + 3 = k(3x^2 + 4x + 2)$.Rearranging the equation:$3(k - 1)x^2 + 4(k - 1)x + (2k - 3) = 0$.For real roots,the discriminant $D \geq 0$:$D = [4(k - 1)]^2 - 4[3(k - 1)][2k - 3] \geq 0$$16(k - 1)^2 - 12(k - 1)(2k - 3) \geq 0$$4(k - 1) [4(k - 1) - 3(2k - 3)] \geq 0$$4(k - 1) [4k - 4 - 6k + 9] \geq 0$$4(k - 1)(-2k + 5) \geq 0$$(k - 1)(2k - 5) \leq 0$.Solving the inequality,we get $1 \leq k \leq \frac{5}{2}$.However,if $k = 1$,the equation becomes $0x^2 + 0x - 1 = 0$,which has no solution. Thus,$k 
eq 1$.Therefore,$k \in (1, \frac{5}{2}]$.

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:

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