Prove that the set of value(s) of $a$ for which the points $P(u,-w)$ and $Q(v,a^2)$ lies on the same side of the line $4x-y+5=0$ is $(-3,3).$

Question

Let $u,v,w$ satisfy the equations $uvw=-6,uv+vw+wu=-5,u+v+w=2,$where $u>v>w$,then prove that the set of value(s) of $a$ for which the points $P(u,-w)$ and $Q(v,a^2)$ lies on the same side of the line $4x-y+5=0$ is $(-3,3).$

---

I tried to solve it.Since we know that when two points $(x_1,y_1),(x_2,y_2)$ lie on the same side of the line $ax+by+c=0$,then $(ax_1+by_1+c)(ax_2+by_2+c)>0$.So i applied this here.

$(4u+w+5)(4v-a^2+5)>0$

$16uv-4ua^2+20u+4vw-a^2w+5w+20v-5a^2+25>0$
But i dont know how to further simplify it to get the answer.Please help me.Thanks.

Answer 1

First we will calculate values of $u,v,w\;\;,$ Where $u>v>w$

Given $$u+v+w=2\;\;,uv+vw+wu=-5\;\;,uvw=-6\;\;$$

Now Let we will form a cubic equation in terms of $x$ whose roots are $x=u\;,x=;,x=w$

So Using factor theorem $(x-u)\;\;,(x-v)\;\;,(x-w)$ are three factors of cubic expression.

So $$(x-u)(x-v)(x-w) = 0\Rightarrow x^3-(u+v+w)x^2+(uv+vw+wu)x-uvw=0$$

So $$x^3-2x^2-5x+6=0$$

Clearly $x=1$ satisfy the above equation. So $(x-1)$ is a factor of above cubic expression.

So $$(x-1)(x-3)(x+2) = 0\Rightarrow x=-2\;\;,x=1\;\;,x=3$$

So we get $u=3\;\;,v=1\;\;,w=-2$

Now Using what you have done in your process.

Since we know that when two points $(x_1,y_1),(x_2,y_2)$ lie on the same side of the line

$ax+by+c=0$,then $(ax_1+by_1+c)(ax_2+by_2+c)>0$.

So i applied this here.

$$(4u+w+5)(4v-a^2+5)>0\Rightarrow 15\cdot (4-a^2+5)>0$$

So we get $$(9-a^2)>0\Rightarrow a^2-9<0\Rightarrow (a+3)(a-3)<0$$

So we get $a\in \left(-3,3\right)$