Gven a,b,c>0 and a+b+c=1 and equation ax^2+bx+c=x have a root 0<x0<1,pls show 2a+b>1?

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The problem provides the information that `x_0` is a solution to equation `ax^2 + bx + c = x` , hence, you need to replace `x_0` for x in equation, such that:

`ax_0^2 + bx_0 + c = x_0`

Since the problem provides the information that `a + b + c = 1` , hence, you may rewrite the equation such that:

`ax_0^2 + bx_0 + c = (a + b + c)x_0`

`ax_0^2 + bx_0 + c = ax_0 + bx_0 + cx_0`

Reducing duplicate members yields:

`ax_0^2 + c = ax_0 + cx_0`

You need to move all terms to one side, such that:

`ax_0^2 + c - ax_0 - cx_0 = 0`

Grouping the terms yields:

`(ax_0^2 - ax_0) - (c + cx_0) = 0`

`ax_0(x_0 - 1) - c(x_0 - 1) = 0`

Factoring out `(x_0 - 1)` yields:

`(x_0 - 1)(ax_0 - c) = 0`

Using zero produt property yields:

`{(x_0 - 1 = 0),(ax_0 - c = 0):}`

Since the equation `x_0 - 1 = 0` is invalid because `x_0 < 1` yields that only equation `ax_0 - c = 0` holds, such that:

`ax_0 - c = 0 => ax_0 = c => x_0 = c/a`

Since `0 < x_0 < 1` yields:

`0 < c/a < 1 => c < a`

Using the condition `a + b + c = 1` yields:

`a + b + c = 1 < a + b + a = 2a + b`

**Hence, evaluating if the inequality `2a + b > 1` holds, under the given conditions, yields that the statement` 2a + b > 1` is valid.**

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