Solve 2x^2=3x+2 by graphing. If excat roots cannot be found, state the consecutive integers between which the roots are located.
I don't have an answer but I made a table and plugged in these values for x respectfully: -1/4, -2/4, -3/4, -1, 0
Thats all I have
somehow it's hard to find the roots manually since
b^2 - 4ac <0
first of all bring all terms to one side
2x^2 - 3x - 2 = 0
now let f(x) = 2x^2 - 3x - 2
did you notice that it factors?
f(x) = (2x+1)(x-2)
so the x-intercepts are -1/2 and 2
which then become the roots of the equation from above.
To graphically solve the equation 2x^2 = 3x + 2, you can start by rearranging it into the standard quadratic form ax^2 + bx + c = 0, where a = 2, b = -3, and c = -2.
1. Plot the quadratic function: Draw the graph of the quadratic function y = 2x^2 - 3x - 2. You can choose a range of x-values, such as -3 to 3, and calculate the corresponding y-values for each x-value using the equation.
2. Identify the x-intercepts: On the graph, locate the x-values where the function intersects or crosses the x-axis. These are the values of x for which y is equal to zero. In other words, find the values of x where 2x^2 - 3x - 2 = 0.
3. Determine the roots: The x-intercepts on the graph represent the roots of the equation. If you find exact points of interception, those are the exact roots. If not, you can determine the consecutive integers between which the roots are located by analyzing the intervals on the x-axis where the graph crosses or touches the x-axis.
To improve your table method:
1. Choose a range of values for x that includes the suspected x-intercepts based on the graph.
2. Plug in each x-value from your chosen range into the equation 2x^2 - 3x - 2.
3. Calculate the corresponding y-values for each x-value.
4. Note the values of x where the y-values are close to zero or change sign (+ to - or - to +). Between those x-values, the roots of the equation should lie.