# Please calculate (a+2)/(b-4)+(b+2)/(a-4) given a, b the roots of quadratic 2x^2 - 4x -5 = 0?

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### 1 Answer

You need to evaluate the given expression `(a + 2)/(b - 4) + (b + 2)/(a - 4)` that depends on the roots of equation, hence, either you calculate a,b using quadratic formula, or you may use Vieta's formulas.

Using Vieta's formulas, you need to bring the terms of expression to a common denominator, such that:

`(a + 2)/(b - 4) + (b + 2)/(a - 4) = ((a + 2)(a - 4) + (b + 2)(b - 4))/((b - 4)(a - 4))`

`(a + 2)/(b - 4) + (b + 2)/(a - 4) = (a^2 - 2a - 8 + b^2 - 2b - 8)/(ab - 4(a + b) + 16)`

Since a quadratic equation holds if you replace the roots into equation, you may use this aspect in the following manner, such that:

`2a^2 - 4a - 5 = 0`

`2b^2 - 4b - 5 = 0`

You may multiplicate the numerator and denominator of expression you need to evaluate by `2` , such that:

`(2a^2 - 4a - 16 + 2b^2 - 4b - 16)/(2ab - 8(a + b) + 32)`

Using quadratic relations `2a^2 - 4a - 5 = 0` and `2b^2 - 4b - 5 = 0` yields:

`2a^2 - 4a - 5 = 0 => 2a^2 - 4a = 5`

`2b^2 - 4b - 5 = 0 => 2b^2 - 4b = 5`

Replacing `5` for `2a^2 - 4a` and `2b^2 - 4b` in expression, yields:

`(5 - 16 + 5 - 16)/(2ab - 8(a + b) + 32) = -22/(2ab - 8(a + b) + 32)`

You need to use Vieta's relations, such that:

`a + b = -(-4/2) = 2`

`a*b = -5/2`

Replacing `2` for` a + b` and `-5/2 ` for `ab` yields:

`-22/(2ab - 8(a + b) + 32) = -22/(2(-5/2) - 8*2 + 32)`

`-22/(2ab - 8(a + b) + 32) = -22/(-5 - 16 + 32)`

`-22/(2ab - 8(a + b) + 32) = -22/11 = -2`

**Hence, evaluating the given expression `(a + 2)/(b - 4) + (b + 2)/(a - 4)` , using Vieta's formulas, yields `(a + 2)/(b - 4) + (b + 2)/(a - 4) = -2` ** .

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