Krdytkyu

The equation
$$sqrt{9-4x}=p-2x$$
has exactly 2 real and different solutions only if parameter $p$ belongs to which set?

So what I see here, to have the solutions be real in the first place,

$9-4xge0$

So $-4xge-9 Rightarrow xle{9over4}$

$xinleft(-infty,{9over4}right]$

From here I quadrate both sides so I get 2 different possibilities:

1. For $pge2x$:

$9-4x=p^2-4x+4x^2\9-4x-p^2+4x-4x^2=0\-4x^2-p^2+9=0$

And for this to have 2 real and different solutions, $D>0$

$-4(-4)(-p^2+9)>0\16(-p^2+9)>0\-16p^2+144>0\-16p^2>-144\p^2<9\|p|<3 Rightarrow pin(-3,3)$

2. For $p<2x:$

$9-4x=-p^2+4x-4x^2\9-4x+p^2+4x^2-4x=0\4x^2-8x+p^2+9=0$

Again $D>0$

$(-8)^2-4cdot4(p^2+9)>0\64-16p^2-144>0\p^2<-5\pnotinBbb R$

The problem here is I don't know how to use $xinleft(-infty,{9over4}right]$ in both cases, so I don't exactly know the solution to this problem. From the textbook I got this question from the result is $pinleft[{9over2},5right)$

As already pointed out in a comment, $(p-2x)^2 = p^2-4px+4x^2$, which results in
$$4x^2 - 4(p-1)x+(p^2-9) = 0,$$
or
$$x = frac{(p-1)pmsqrt{10-2p}}{2}.$$
For two real distinct roots, $10-2p > 0 implies color{red}{p < 5}$.

Also, $p ge 2x$ implies $$(p-1)pmsqrt{10-2p} le p implies sqrt{10-2p}le 1 implies color{red}{p ge frac{9}{2}}.$$

Note that