# Given equation x2+mx+8=0, find quadratic with roots y1=x1+x2,y2=x1x2?

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### 1 Answer

You should use Vieta's formulas for a quadratic equation `ax^2 + bx + c = 0` such that:

`x_1 + x_2 = -b/a`

`x_1*x_2 = c/a`

You need to identify a,b,c such that:

`x_1 + x_2 = -m/1`

`x_1*x_2 = 8/1`

You need to find `y_1` and `y_2` using the conditions provided by the problem such that:

`y_1 = x_1 + x_2 = -m`

`y_2 = x_1*x_2 = 8`

You may form the quadratic equation using Lagrange's resolvents such that:

`y^2 - (y_1+y_2)*y + (y_1*y_2) = 0`

`y_1+y_2 = 8 - m`

`y_1*y_2 = -8m `

You should substitute 8 - m for `y_1+y_2` and -8m for `y_1*y_2` such that:

`y^2 - (8 - m)*y- 8m= 0`

**Hence, evaluating the quadratic equation under given conditions yields `y^2 - (8 - m)*y - 8m = 0.` **

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