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 Homework Statement:
 Suppose ##x_1## and ##x_2## are roots of the quadratic equation ##x^2+2(k3)x+9=0## ##\left( x_1 \neq x_2 \right)##. For what values of ##k## do the inequalities ##6<x_1<1## and ##6<x_2<1## hold true?
 Relevant Equations:

(1) For real roots of a quadratic equation ##ax^2+bx+c=0##, its discriminant ##\mathscr{D} = b^24ac \geq 0##.
(2) The two roots ##[x_1, x_2]## of the quadratic equation above are equal to ##x_{1,2}= \frac{b\pm \sqrt{b^24ac}}{2a}##. We can have ##x_1## as the larger of the two roots : ##x_1>x_2##. This would imply ##x_1## takes the plus sign ##+## and ##x_2## the minus sign #### in the expression for the roots.
Given equation and conditions: ##\boldsymbol{x^2+2(k3)x+9=0}##, with roots ##\boldsymbol{(x_1,x_2)}##. These roots satisfy the condition ##\boldsymbol{6<x_1,x_2<1}##.
Question : ##\text{What are the allowable values for}\; \boldsymbol{k}?##
(0) Let me take care of the determinant first, for real roots. The determinant ##\mathscr {D} = 4(k3)^24\times 9 \geq 0\Rightarrow (k3)^2>9## which would imply that either ##k3 \geq 3## or ##k3 \leq 3##. These lead to ##\boxed{k \geq 6\; \text {OR}\; k \leq 0}##.
(1) Let us have the equation again : ##x^2+2(k3)x+9=0##. Bigger root, as per the given condition, ##x_1 = \frac{2(k3)+\sqrt{\mathscr{D}}}{2}<1##. Now since the discriminant ##\mathscr{D} \geq 0##, this implies ##(k3)<1\Rightarrow k3 >1 \Rightarrow k >4##. But from point (0) above, we must have ##k \geq 6##. Hence, taking the stronger of the two requirements, this condition yields ##\boxed{k\geq 6}##.
Now here's the interesting bit. The bigger root ##x_1## also has a lower desired limit as per the problem, ##6##. However, so does the smaller root ##x_2##. If I can show the cases of ##k## for which the smaller root ##x_2 > 6##, clearly for those values, the larger root will also be ##> 6##. I hope my reasoning is sound, but it most likely isn't, for it gives me the wrong answer as I will quote below.
(2) Now the smaller root ##x_2 = \frac{2(k3)\sqrt{\mathscr{D}}}{2}>6##. Since the discriminant ##\mathscr{D} \geq 0##, it implies that ##(k3)> 6\Rightarrow k3 < 6 \Rightarrow \boxed{k<3}##. We find that this result is stronger than the one in point (0) earlier, so we stick with it.
Hence, my answers are, putting the results of (0), (1) and (2) : ##\boxed{k \geq 6\; \text{OR} \; k<3}##.
Answer from the book : ##\boxed{x \in (6;6.75)}## ##\Rightarrow 6 \le x \le 6.75##.
A help would be welcome.
Question : ##\text{What are the allowable values for}\; \boldsymbol{k}?##
(0) Let me take care of the determinant first, for real roots. The determinant ##\mathscr {D} = 4(k3)^24\times 9 \geq 0\Rightarrow (k3)^2>9## which would imply that either ##k3 \geq 3## or ##k3 \leq 3##. These lead to ##\boxed{k \geq 6\; \text {OR}\; k \leq 0}##.
(1) Let us have the equation again : ##x^2+2(k3)x+9=0##. Bigger root, as per the given condition, ##x_1 = \frac{2(k3)+\sqrt{\mathscr{D}}}{2}<1##. Now since the discriminant ##\mathscr{D} \geq 0##, this implies ##(k3)<1\Rightarrow k3 >1 \Rightarrow k >4##. But from point (0) above, we must have ##k \geq 6##. Hence, taking the stronger of the two requirements, this condition yields ##\boxed{k\geq 6}##.
Now here's the interesting bit. The bigger root ##x_1## also has a lower desired limit as per the problem, ##6##. However, so does the smaller root ##x_2##. If I can show the cases of ##k## for which the smaller root ##x_2 > 6##, clearly for those values, the larger root will also be ##> 6##. I hope my reasoning is sound, but it most likely isn't, for it gives me the wrong answer as I will quote below.
(2) Now the smaller root ##x_2 = \frac{2(k3)\sqrt{\mathscr{D}}}{2}>6##. Since the discriminant ##\mathscr{D} \geq 0##, it implies that ##(k3)> 6\Rightarrow k3 < 6 \Rightarrow \boxed{k<3}##. We find that this result is stronger than the one in point (0) earlier, so we stick with it.
Hence, my answers are, putting the results of (0), (1) and (2) : ##\boxed{k \geq 6\; \text{OR} \; k<3}##.
Answer from the book : ##\boxed{x \in (6;6.75)}## ##\Rightarrow 6 \le x \le 6.75##.
A help would be welcome.