# If a and b are the roots of equation 2x^2-4x+6=0,form the quadratic equation with the roots(a+2b) and (b+2a).

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You should use Vieta's relations such that:

`x_1 + x_2 = -(-4)/2`

`x_1 + x_2 = 2`

`x_1*x_2 = 6/2 =gt x_1*x_2 = 3`

The problem provides the information that the roots of the quadratic equation are a and b, hence, you should substitute a and b for `x_1` and `x_2` such that:

`a+b = 2`

`a*b = 3`

You should form the quadratic equation that has the roots a+2b and b+2a, hence, you should use Lagrange's resolvents such that:

`x^2 - px + q = 0`

`p = a + 2b + b + 2a`

`p = 3(a+b)`

Notice that from Vieta's relations, you may substitute 2 for a+b such that:

`p = 3*2 = 6`

You need to find q such that:

`q = (a+2b)(b+2a)`

`q = ab + 2a^2 + 2b^2 + 4ab`

`q = 5ab + 2(a^2+b^2)`

You need to evaluate `a^2+b^2` such that:

`a^2+b^2 = (a+b)^2 - 2ab`

Substituting 2 for a+b and 3 for ab yields:

`a^2+b^2 = (2)^2 - 2*3`

`a^2+b^2 = 4 - 6`

`a^2+b^2 = 2`

You should substitute 2 for `a^2+b^2` in expression of q such that:

`q = 5*3+ 2*2 =gt q = 15 + 4`

`q = 19`

Hence, substituting 6 for p and 19 for q yields:

`x^2 - 6x + 19 = 0`

**Hence, evaluating the quadratic equation under the given conditions yields `x^2 - 6x + 19 = 0.` **

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