- #1

brotherbobby

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- Homework Statement
- Suppose ##x_1## and ##x_2## are roots of the quadratic equation ##x^2+2(k-3)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^2-4ac \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^2-4ac}}{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(k-3)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(k-3)^2-4\times 9 \geq 0\Rightarrow (k-3)^2>9## which would imply that either ##k-3 \geq 3## or ##k-3 \leq -3##. These lead to ##\boxed{k \geq 6\; \text {OR}\; k \leq 0}##.

(1) Let us have the equation again : ##x^2+2(k-3)x+9=0##. Bigger root, as per the given condition, ##x_1 = \frac{-2(k-3)+\sqrt{\mathscr{D}}}{2}<1##. Now since the discriminant ##\mathscr{D} \geq 0##, this implies ##-(k-3)<1\Rightarrow k-3 >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(k-3)-\sqrt{\mathscr{D}}}{2}>-6##. Since the discriminant ##\mathscr{D} \geq 0##, it implies that ##-(k-3)> -6\Rightarrow k-3 < -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.**