For any quadratic equation, if you have irrational or complex roots, they will always come as conjugate pairs, just like your given roots
method one:
use the fact that if x=a is a root, then x-a will be a factor of your equation, so
1.
(x - (2+√5))(x - (2-√5) = 0
(x-2 - √5)(x - 2 + √5) = 0
x^2 - 2x + √5x - 2x + 4 - 2√5 - √5x + 2√5 - 5 = 0
x^2 - 4x - 1 = 0
method 2:
use the property that for x^2 + bx + c = 0
the sum of the roots is -b and the product of the roots is c
1. sum of roots = 2+√5 + 2-√5 = 4
product of the roots = (2+√5)(2-√5) = 4 - 5 -1
so the equation is
x^2 -4x - 1 = 0
use either method for your other two equations.
check by finding their roots.
Write the quadratic function whose zeros are:
1. 2+square root of 5, 2-square root of 5
2. 3+7i,3-7i
3. 5±i
Please tell me how to get them. I have no clue on how to solve them.
1 answer