Construct the quadratic that has one root (1/3)+square root (2/3).
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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.