Find the polynomials roots to each of the following problems:
#1) x^2+3x+1
#2) x^2+4x+3=0
#3) -2x^2+4x-5
To find the roots of a quadratic polynomial, we can use the quadratic formula. The quadratic formula states that for a quadratic polynomial of the form ax^2 + bx + c = 0, the roots (also known as solutions) can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Let's use this formula to find the roots of each problem:
#1) x^2 + 3x + 1 = 0
Here, a = 1, b = 3, and c = 1. Plugging these values into the quadratic formula, we get:
x = (-3 ± √(3^2 - 4(1)(1))) / 2(1)
x = (-3 ± √(9 - 4)) / 2
x = (-3 ± √5) / 2
So the roots of the polynomial are (-3 + √5)/2 and (-3 - √5) / 2.
#2) x^2 + 4x + 3 = 0
Here, a = 1, b = 4, and c = 3. Plugging these values into the quadratic formula, we get:
x = (-4 ± √(4^2 - 4(1)(3))) / 2(1)
x = (-4 ± √(16 - 12)) / 2
x = (-4 ± √4) / 2
x = (-4 ± 2) / 2
Simplifying, we get x = -1 and x = -3 as the roots of the polynomial.
#3) -2x^2 + 4x - 5 = 0
Here, a = -2, b = 4, and c = -5. Plugging these values into the quadratic formula, we get:
x = (-4 ± √(4^2 - 4(-2)(-5))) / 2(-2)
x = (-4 ± √(16 - 40)) / -4
x = (-4 ± √(-24)) / -4
Since we have a negative value under the square root, this means the polynomial has no real roots. The roots would be complex numbers, but we haven't discussed how to find complex roots in this explanation.