Question

Write the quadratic equation that has roots -1-rt2/3 and -1+rt2/3 if its coefficient with x^2 is equal to 1

Answer 1

The equation of the quadratic function is f(x) = x²+ 2/3x - 1/9

How to determine the quadratic equation?

From the question, the given parameters are:

Roots = (-1 - √2)/3 and (-1 + √2)/3

The quadratic equation is then calculated as

f(x) = The products of (x - roots)

Substitute the known values in the above equation

So, we have the following equation

$f(x) = (x - \frac{-1-\sqrt{2}}{3})(x - \frac{-1+\sqrt{2}}{3})$

This gives

$f(x) = (x + \frac{1+\sqrt{2}}{3})(x + \frac{1-\sqrt{2}}{3})$

Evaluate the products

$f(x) = (x^2 + \frac{1+\sqrt{2}}{3}x + \frac{1-\sqrt{2}}{3}x + (\frac{1-\sqrt{2}}{3})(\frac{1+\sqrt{2}}{3})$

Evaluate the like terms

$f(x) = x^2 + \frac{2}{3}x - \frac{1}{9}$

So, we have

f(x) = x²+ 2/3x - 1/9

Read more about quadratic equations at

brainly.com/question/1214333

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