- #1

- 3,884

- 11,585

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- #1

- 3,884

- 11,585

- #2

- 3,884

- 11,585

Rewrite $g(x)$ in terms of $a$ and $b$...

- #3

- 3,884

- 11,585

$=a=r+s+t,\,b=rs+st+tr,\,1=rst$

and

$m=-(r^2+s^2+t^2),\,n=r^2s^2+s^2t^2+t^2r^2,\,p=-r^2s^2t^2$

Let's try to express $m,\,n$ and $p$ in terms of $a$ and $b$. The easiest one is $p$:

$p=-(rst)^2=-1$

From $m$, we square $r+s+t$:

$a^2=(r+s+t)^2=r^2+s^2+t^2+2(rs+st+tr)=-m+2b\implies m=2b-a^2$

Finally, for $n$ we can square $rs+st+tr$:

$b^2=(rs+st+tr)^2=r^2s^2+s^2t^2+t^2r^2+2(r^2st+rs^2t+t^2rs)=n+2rst(r+s+t)=n-2a\implies n=b^2+2a$

So now we write $g(x)$ in terms of $a$ and $b$:

$g(x)=x^3+(2b-a^2)x^2+(b^2+2a)x-1$

We know that $g(-1)=-5$; when we plug this into our equation for $g(x)$ we get

$-5=(-1)^3+(2b-a^2)(-1)^2+(b^2+2a)(-1)-1=1b-a^2-b^2-2a-2\implies a^2+2a+b^2-2b-3=0$

We seek the largest possible value of $b$. Since $a$ is real, we know that the discriminant of this quadratic must be non-negative. In particular,

$2^2-4(b^2-2b-3)\ge 0\implies b^2-2b-4\le 0$

Solving this quadratic gives us that the largest possible value of $b$ is $1+\sqrt{5}$.

Share:

- Replies
- 1

- Views
- 280

- Replies
- 4

- Views
- 756

- Replies
- 1

- Views
- 561

- Replies
- 2

- Views
- 586

- Replies
- 5

- Views
- 655

- Replies
- 1

- Views
- 553

- Replies
- 1

- Views
- 675

- Replies
- 1

- Views
- 822

- Replies
- 2

- Views
- 613

- Replies
- 3

- Views
- 730