solve 2x^2 + 4x +2 =0 by graphing the corresponding function and determing the zeros

Answer 1

To solve the equation 2x^2 + 4x + 2 = 0, we can graph the corresponding function y = 2x^2 + 4x + 2 and determine the zeros.

First, let's rearrange the equation to get it in the form of y = 0:
2x^2 + 4x + 2 = 0
Subtracting 2 from both sides:
2x^2 + 4x = -2
Dividing the entire equation by 2:
x^2 + 2x = -1

Now, we can graph the function y = x^2 + 2x - 1 and find the zeros using the graph:

The graph of y = x^2 + 2x - 1 is a parabola that opens upwards. Since the coefficient of x^2 is positive (1), the graph will open upwards. To find the zeros, we look for the x-intercepts where the graph crosses the x-axis.

Here is the graph:

Since we are trying to find where y = 0, we need to find the x-values where the graph crosses the x-axis.

By observing the graph, we see that there are two x-intercepts near -2 and 0. To be more precise, we can use the quadratic formula to calculate the actual zeros.

The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = 1, b = 2, and c = -1.

Using the quadratic formula:
x = (-2 ± √(2^2 - 4*1*(-1))) / (2*1)
x = (-2 ± √(4 + 4)) / 2
x = (-2 ± √(8)) / 2
x = (-2 ± 2√2) / 2
x = -1 ± √2

So, the zeros of the equation 2x^2 + 4x + 2 = 0 are approximately:
x ≈ -1 + √2 and x ≈ -1 - √2