# Construct the quadratic that has one root (1/3)+square root (2/3).

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### 2 Answers

The roots of a quadratic equation, if irrational, are found as pairs. Here we have one roots as 1/3 + sqrt 2/3, the other root will be 1/3 - sqrt 2/3

We can determine the quadratic equation by multiplying the factors,

(x - 1/3 - sqrt 2/3)(x - 1/3 + sqrt 2/3)

=> (x - 1/3)^3 - (sqrt 2/3)^2

=> x^2 + 1/9 - 2x/3 - 2/3

=> x^2 - 2x/3 - 5/9

multiply all terms by 9

=> 9x^2 - 6x - 5 = 0

**The required quadratic equation is 9x^2 - 6x - 5 = 0**

If one root of a quadratic is (1+sqrt2)/3, then it's conjugate, (1-sqrt2)/3, is also a root for the quadratic.

We'll use the formula of a quadratic:

x^2 - Sx + P = 0

S is the sum of the roots.

P is the product of the roots.

Since we know both roots, we'll determine their sum and product to create the quadratic.

x1 + x2 = S = 1/3 + (sqrt2)/3 + 1/3 - (sqrt2)/3

We'll eliminate like terms:

x1 + x2 = S = 2/3

The product is:

P = x1*x2 = (1+sqrt2)(1-sqrt2)/9

The numerator represents the difference of 2 squares:

P = (1-2)/9

P = -1/9

We'll create the quadratic: x^2 - 2x/3 - 1/9 = 0

We'll multiply by 9 the quadratic:

9x^2 - 6x - 1 = 0

**The final form of the quadratic is: 9x^2 - 6x - 1 = 0.**