If tan `prop` &tan `beta` are the roots of the equation `bx^2+ax-1=0,bcancel=0` and `tan(prop+beta)=2` then `a+2b+2` is equal to ....

Explanation

To solve the problem, we start with the given quadratic equation:

\[ bx^2 + ax - 1 = 0 \]

where \( \tan \alpha \) and \( \tan \beta \) are the roots. We know from Vieta's formulas that:

1. The sum of the roots \( \tan \alpha + \tan \beta = -\frac{a}{b} \)
2. The product of the roots \( \tan \alpha \tan \beta = -\frac{-1}{b} = \frac{1}{b} \)

We are also given that:

\[ \tan(\alpha + \beta) = 2 \]

Using the tangent addition formula:

\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]

Substituting the values we found from Vieta's formulas, we have:

\[ \tan(\alpha + \beta) = \frac{-\frac{a}{b}}{1 - \frac{1}{b}} = \frac{-\frac{a}{b}}{\frac{b-1}{b}} = \frac{-a}{b-1} \]

Setting this equal to 2 gives us:

\[ \frac{-a}{b-1} = 2 \]

Cross-multiplying yields:

\[ -a = 2(b - 1) \]

Expanding this gives:

\[ -a = 2b - 2 \]

Rearranging gives:

\[ a + 2b - 2 = 0 \]

Thus, we can express \( a + 2b + 2 \) as:

\[ a + 2b + 2 = 0 + 4 = 4 \]

So, the value of \( a + 2b + 2 \) is:

\[ \boxed{4} \]