# Let alpha,beta be the roots of 2x^2-15x+4=0,then the equation whose roots are alpha+3,beta+3 ?

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You need to remember how to use the roots to form a quadratic equation such that:

`x^2 - sx + p = 0`

Since s expresses the sum of roots and p expresses the product of roots, hence you may form the equation that has the roots `alpha + 3 ` and `beta + 3` such that:

`x^2 - (alpha + 3 + beta + 3)x + (alpha + 3)(beta + 3) = 0`

You need to open the brackets such that:

`x^2 - (alpha + beta + 6)x + (alpha*beta + 3(alpha + beta) + 9) = 0`

You need to use Vieta's formulas for the first equation `2x^2-15x+4 = 0` to relate the coefficients of quadratic to its sum and product of roots such that:

`alpha + beta = 15/2`

`alpha*beta = 4/2 = 2`

You should substitute `15/2` for `alpha + beta ` and 2 for `alpha*beta ` in equation `x^2 - (alpha + beta + 6)x + (alpha*beta + 3(alpha + beta) + 9) = 0` such that:

`x^2 - (15/2 + 6)x + (2 + 3(15/2) + 9) = 0`

`x^2 - 27/2x + 67/2 = 0`

`2x^2 - 27x + 67 = 0`

**Hence, evaluating the quadratic equation that has the roots `alpha + 3` and `beta + 3` , under given conditions, yields `2x^2 - 27x + 67 = 0.` **