Finding the range of values of a parameter to satisfy the given condition.

Question

Find the range of values of a for which the one of the roots of the equation:

$(2a+1)x^2-ax+a-2=0$ is greater than and the other smaller than unity (i.e. $1$).

Now, $\Delta>0$ because there are two distinct roots.

Thus, $a \in \left(\dfrac{6-2\sqrt23}{7},\dfrac{6+2\sqrt23}{7}\right) $

I think I will get another range of $a$ and then I'll have to intersect the two sets. But how do I continue from here?

Answer 1

Let $2a+1>0$, i.e. $a>-\dfrac12$.

Then, $f(1)<0$: $(2a+1)-a+a-2=2a-1<0$, i.e. $a<\dfrac12$.

We obtain $-\dfrac12<a<\dfrac12$.

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Let $2a+1<0$, i.e. $a<-\dfrac12$.

Then, $f(1)>0$: $(2a+1)-a+a-2=2a-1>0$, i.e. $a>\dfrac12$.

Contradiction.

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In conclusion, $-\dfrac12<a<\dfrac12$.

Answer 2

Hint: Simply using the quadratic roots formula, you have two roots expressing in 'a'; simply set the larger one be > 1 and smaller one be <1; then you will be a range of 'a' for each of the 2 equations. Considering also your delta condition as you have shown, you would now have 3 ranges; you're right that the final thing to do is the intersect all of them to find a range that satisfy all conditions. Hope it helps.

Answer 3

The domain is: $$D=a^2-4(2a+1)(a-2)\ge 0 \Rightarrow a\in[-\frac27(\sqrt{23}-3)\approx-0.513,\frac27(\sqrt{23}+3) \approx 2.227].$$ Express the equation as: $$(2a+1)x^2=ax-a+2.$$ Consider $f(x)=(2a+1)x^2$ and $g(x)=ax-a+2$. Consider the cases: $$1) \ \begin{cases}2a+1>0 \\ \ \ a \ \ \ \ \ \ \ \ge 0 \end{cases}\Rightarrow \begin{cases}a>-\frac12 \\ a \ge 0 \end{cases} \Rightarrow a\ge 0.$$ $$f(1)<g(1) \Rightarrow 2a+1<2 \Rightarrow a<\frac12$$ Hence: $0\le a<\frac12$. $$2) \ \begin{cases}2a+1>0 \\ \ \ a \ \ \ \ \ \ \ <0 \end{cases}\Rightarrow \begin{cases}a>-\frac12 \\ a <0 \end{cases} \Rightarrow -\frac12<a<0.$$ $$f(1)<g(1) \Rightarrow 2a+1<2 \Rightarrow a<\frac12$$ Hence: $-\frac12<a<0$. $$3) \ \begin{cases}2a+1<0 \\ \ \ a \ \ \ \ \ \ \ \ge 0 \end{cases}\Rightarrow \begin{cases}a<-\frac12 \\ a \ge 0 \end{cases} \Rightarrow \emptyset.$$ Hence: $\emptyset$. $$4) \ \begin{cases}2a+1<0 \\ \ \ a \ \ \ \ \ \ \ <0 \end{cases}\Rightarrow \begin{cases}a<-\frac12 \\ a <0 \end{cases} \Rightarrow a<-\frac12.$$ $$f(1)<g(1) \Rightarrow 2a+1<2 \Rightarrow a<\frac12$$ Hence: $a<-\frac12$.

Finally, taking into account the domain, we get: $$a\in [-\frac27(\sqrt{23}-3),-\frac12)\cup (-\frac12,\frac12)$$