Question

How to show that `y>(9/10)` from equation `y=(9+x^2)/(x^2+2x+1)`?

Answer 1

Actually:

`y >= 9/10`

as follows...

Explanation:

Given:

`y = (9+x^2)/(x^2+2x+1)`

Multiply both sides by `x^2+2x+1` to get:

`yx^2+2yx+y = x^2+9`

Subtract `x^2+9` from both sides to get:

`(y-1)x^2+2yx+(y-9) = 0`

This is a quadratic equation in `x` in standard form:

`ax^2+bx+c = 0`

with `a=y-1`, `b=2y` and `c=y-9`

The discriminant `Delta` of this quadratic is given by the formula:

`Delta = b^2-4ac`

`color(white)(Delta) = (2y)^2-4(y-1)(y-9)`

`color(white)(Delta) = 4y^2-4(y^2-10y+9)`

`color(white)(Delta) = 4y^2-4y^2+40y-36`

`color(white)(Delta) = 40(y-9/10)`

In order that the quadratic equation in `x` has a real root, we must have `Delta >= 0`

So:

`40(y-9/10) >= 0`

Divide both sides by `40` to find:

`y-9/10 >= 0`

Add `9/10` to both sides to get:

`y >= 9/10`

Let's find where `y=9/10` ...

`(9/10 - 1)x^2+2(9/10)x+(9/10-9) = 0`

That is:

`-1/10x^2+18/10x-81/10 = 0`

Multiply through by `-10` to get:

`x^2-18x+81 = 0`

That is:

`(x-9)^2 = 0`

So:

`x = 9`

graph{(y-(9+x^2)/(x^2+2x+1))((x-9)^2+(y-9/10)^2-0.03) = 0 [-7.73, 12.27, -3.82, 6.18]}