Are the zeros of g(x) = 2x^2 - 12x + 40 and g(x) = x^2 - 6x + 20 the same?
Yes, becouse:
2 x ^ 2 - 12 x + 40 = 0 Divide both sides by 2
x ^ 2 - 6 x + 20 = 0
First equation is second equation multiplied by 2
and
2 * 0 = 0
To determine if the zeros of two polynomials are the same, we need to find the values of x that make each polynomial equal to zero.
Let's start by finding the zeros of g(x) = 2x^2 - 12x + 40.
To find the zeros, we set g(x) equal to zero and solve for x:
2x^2 - 12x + 40 = 0
This quadratic equation can be solved using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For g(x) = 2x^2 - 12x + 40, the values of a, b, and c are as follows: a = 2, b = -12, and c = 40. Plugging these values into the quadratic formula, we get:
x = (-(-12) ± √((-12)^2 - 4(2)(40))) / (2(2))
Simplifying the expression further:
x = (12 ± √(144 - 320)) / 4
= (12 ± √(-176)) / 4
Since the term inside the square root (√(-176)) is negative, we can conclude that the quadratic equation g(x) = 2x^2 - 12x + 40 has no real solutions. Therefore, it does not have any zeros.
Now let's find the zeros of g(x) = x^2 - 6x + 20.
Setting g(x) equal to zero and solving for x, we have:
x^2 - 6x + 20 = 0
This is another quadratic equation. We can once again use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, the values of a, b, and c for g(x) = x^2 - 6x + 20 are: a = 1, b = -6, and c = 20. Plugging these values into the quadratic formula, we get:
x = (-(-6) ± √((-6)^2 - 4(1)(20))) / (2(1))
= (6 ± √(36 - 80)) / 2
= (6 ± √(-44)) / 2
Since the term inside the square root (√(-44)) is negative, we once again have no real solutions. Hence, g(x) = x^2 - 6x + 20 does not have any zeros.
Therefore, both g(x) = 2x^2 - 12x + 40 and g(x) = x^2 - 6x + 20 have no real zeros, but they do not share the same zeros since they have no zeros in common.