Ytukyg

Find real values of $a$ for which the equation

$4^x-(a-3)cdot 2^x+(a+4)=0$
has non positive roots

Try: Let $2^x=yin (0,1].$ Then equation convert into

$y^2-(a-3)y+(a+4)=0$

For real roots its discriminant $geq 0$

So $$(a-3)^2-4(a+4)geq 0$$

$$a^2-10a-7geq 0$$

$$ain bigg(-infty,5-4sqrt{2}bigg]cup bigg[5+4sqrt{2},inftybigg).$$

but answer is different from that , i did not know where i

am missing. could some help me to solve it. thanks

Rewrite the equation as
$$
a=frac{4^x+3cdot 2^x+4}{2^x-1}
$$
and plot the function $a(x)$. From the plot you can see the values of $a$ that correspond to $x<0$.

You can change $y=2^x$ to simplify drawing, but then you will need to neglect negative $y$ as they do not correspond to real values of $x$.

Differentiate to get two critical points $y=1pm 2sqrt{2}$ among which only $1+2sqrt{2}$ is positive, then study intervals of monotonicity and asymptotics. The answer is $a<-4$.

Hint:

your equation has non positive roots if $0<2^x<1$, so, with your substitution, the condition becomes $0<y<1$, for the equation $y^2-(a-3)y+(a+4)=0$

Can you do from this?

If $y_1,y_2$are the solutions we have :

$
y_1+y_2=a-3 qquad ,qquad y_1y_2=a+4
$

and we want:
$$
begin{cases}
0<a-3<2\
0<a+4<1
end{cases}
$$
but: $a+4<1 Rightarrow a<-3$ and $a-3>0 Rightarrow a>3$

so it is impossible to have $0<y_1<1$ and $0<y_2<1$

You want $f(y) = y^2 - (a-3) y + (a+4)$ to have at least one root in $(0,1]$.
Note that $f(0) = a+4$ and $f(1) = 8$, while the minimum value of $f(y)$, occurring at
$y = (a-3)/2$, is $(-a^2 + 10 a + 7)/4$. There are several cases to consider.

• If $a < -4$, $f(0) < 0 < f(1)$ so there is a root in $(0,1]$.


• For $a = -4$, $f(y) = y^2 + 7 y$ so the roots $0$ and $-7$ are not in $(0,1]$.


• For $a > -4$, $f(0) > 0$ and $f(1) > 0$ so the only way to have a root in $(0,1]$ would be if the minimum occurs between $0$ and $1$ and the minimum value $le 0$. The
  minimum is at $(a-3)/2$, so this requires $a in (3,5]$. But for $a$ in this interval,
  $-a^2 + 10 a + 7 > 0$.

So the answer is $a < -4$.

Alternative method:
For given $y$, $f(y) = 0$ for $a = A(y) = 4 + y + 8/(y-1)$. Thus you want the set of values $A(y)$ for $0 < y < 1$ (of course $A(1)$ is undefined).
$A'(y) = dfrac{y^2 - y - 7}{(y-1)^2} < 0$ in this interval, so $A$ is a decreasing function of $y$, with $A(y) to -infty$ as $y to 1-$. Thus the answer is $(-infty, A(0)) = (-infty, -4)$.

Not only the equation in $y$ must have real roots, but these roots have to be $le 1$, since it's asked that the equation in $x$ has non-positive roots, so that $y=2^xle 2^0=1$.

To test whether the roots are less than $1$, we can place $1$ w.r.t. the roots $y_1,y_2$ of $;p(y)=y^2-(a-3)y+a+4$. The standard method is to determine the sign of $p(1)=8>0$. Therefore $1$ does not separate the roots of $p(y)$, and $1$ is either less than or greater than both of them. Furthermore, the half-sum of the roots is $;frac{a-3}2$ by Vieta's relations, so that
$$y_1, y_2 le 1iff frac{y_1+y_2}2=frac{a-3}2 le 1iff ale 5. $$
So the solution is
$$ain Bigl{Bigl(-infty,5-4sqrt{2}Bigr]cup Bigl[5+4sqrt{2},inftyBigr)Bigr}cap (-infty,5]=Bigl(-infty,5-4sqrt{2}Bigr].$$