Find all the values of the parameter $c$ for which the inequality has atleast one solution.

Question

Find all the values of the parameter $c$ for which the inequality has atleast one solution.
$$1+\log_2(2x^2+2x+\frac{7}{2})\geq\log_2(cx^2+c)$$

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First i checked the domain of the inequality,
$2x^2+2x+\frac{7}{2}>0$ and $cx^2+c>0$
First inequality will always be positive for $x\in R$ because its discriminant is negative.And second inequality will be positive provided $c$ is positive.
Then i simplified the given inequality.

$$1+\log_2(2x^2+2x+\frac{7}{2})\geq\log_2(cx^2+c)$$
$4x^2+4x+7\geq cx^2+c$

$(4-c)x^2+4 x+(7-c)\geq 0$
In order to have no solution of the above inequality,we should have discriminant $\geq 0$ and in order to have atleast one solution of the above inequality,we should have discriminant $<0$

$16-4(4-c)(7-c)<0$
$c^2-11c+24>0$ gives me $c\in (-\infty,3)\cup(8,\infty)$

But the answer given is $0<c\leq 8$.I dont know where i am wrong?

Answer 1

Firstly we require $c>0$ for the second log to be real.

Now considering your inequality $$(4-c)x^2+4x+(7-c)\geq0$$

If $c<4$ there will always be at least one solution to the inequality aince the curve is a positive quadratic.

If $c>4$, we require that the discriminant of this negative quadratic to be positive or equal to zero so that the curve does not lie entirely below the $x$ axis.

Therefore $$16-4(4-c)(7-c)\geq0$$ $$\Rightarrow (c-3)(c-8)\leq 0$$ $$\Rightarrow 3\leq c\leq8$$ Hence the requirement for $c$, when we combine these results, is $$0<c\leq8$$

Answer 2

No for one solution discriminant has to be equal to $0$ and for two solutions its $<,>0$ as $log(z)=log(x)+i\theta$ where $|z|=x,\theta$ is its argument. If its real solutions then discriminant is $>0$. Atleast one means its parabola with vertex at $X-axis$ and two means parabola witg two points of intersection on X-axis. Answer to your comment $(c-3)(c-8)>0$ so c(0,8]$