When roots of polynomials occur in radical form, they occur as two conjugates.
That is,
The conjugate of (a + √b) is (a - √b) and vice versa.
To show that the given conjugates come from a polynomial, we should create the polynomial from the given factors.
The first factor is x - (a + √b).
The second factor is x - (a - √b).
The polynomial is
f(x) = [x - (a + √b)]*[x - (a - √b)]
= x² - x(a - √b) - x(a + √b) + (a + √b)(a - √b)
= x² - 2ax + x√b - x√b + a² - b
= x² - 2ax + a² - b
This is a quadratic polynomial, as expected.
If you solve the quadratic equation x² - 2ax + a² - b = 0 with the quadratic formula, it should yield the pair of conjugate radical roots.
x = (1/2) [ 2a +/- √(4a² - 4(a² - b)]
= a +/- (1/2)*√(4b)
= a +/- √b
x = a + √b, or x = a - √b, as expected.
Answer:
100
Step-by-step explanation:
they are vertical angles therefore, they have the same measure.
Answer: Bazillions
Step-by-step explanation: it is bazillions affter zillions
2/3 - 1/4 ... let's find the common numbers 12 so new numbers would be 8/12 and 3/12 aka 11/12 + .. anyways it's 11/12 plus 2/3 x
Answer:
It's 2b/3
Step-by-step explanation:
