- #1
brotherbobby
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- Homework Statement
- Find all the values of ##m## for which both roots of the equation ##2x^2+mx+m^2-5 = 0##
(a) are less than ##1##;
(b) exceed ##-1##.
- Relevant Equations
- For a quadratic equation ##ax^2+bx+c=0## :
(1) For roots to be real, the discriminant ##\mathscr{D} = b^2-4ac\ge0##.
(2) The sum of the roots ##S = -\frac{b}{a}##.
(3) The product of the roots ##P = \frac{c}{a}##.
Given equation : ##\boldsymbol{2x^2+mx+m^2-5 = 0}##.
For the roots of this equation to be real, the discriminant : ##m^2-8(m^2-5) \ge 0\Rightarrow 7m^2-40\le 0\Rightarrow -\sqrt{\frac{40}{7}} \le m \le \sqrt{\frac{40}{7}}##. (This is the same as##\boldsymbol{-2.39\le m \le 2.39}##, in case if actual amounts become necessary as "cut offs" from later constraint equations of a similar kind).
(a) If the roots are ##\alpha, \beta##, the first given condition is that ##\alpha, \beta < 1## ##\Rightarrow \alpha \beta < 1##. This amounts to ##\frac{m^2-5}{2} < 1\Rightarrow m^2 < 7 \Rightarrow -\sqrt{7} < m < \sqrt{7}## which is the same as ##-2.65 < m < 2.65##. Since the earlier constraint for real roots (see above in bold) is more restrictive than this, my answer for this case : ##\boxed {\sqrt{\frac{40}{7}} \le m \le \sqrt{\frac{40}{7}}}##.
(b) For the second condition, ##\alpha, \beta > -1## ##\Rightarrow \alpha + \beta > -1 \Rightarrow -\frac{m}{2} > -1 \Rightarrow m < 2## . "Uniting" this with the solution which demands both roots be real, my answer to this part is ##\boxed{-\sqrt{\frac{40}{7}} \le m < 2}##.
However, according to the book, both my answers are incorrect.
The answers according to the text are : (a) ## m \in \left[ -\sqrt{\frac{40}{7}} ; - \frac{1+\sqrt{13}}{2}\right ) \cup \left(\frac{\sqrt{13} - 1}{2}; \sqrt{\frac{40}{7}} \right ]## and (b) ## m \in \left[ -\sqrt{\frac{40}{7}} ; \frac{1-\sqrt{13}}{2}\right ) \cup \left(\frac{\sqrt{13} + 1}{2}; \sqrt{\frac{40}{7}} \right ]##.
Let me copy and paste the answers as they appear.
Any help would be appreciated.