Solving quadratic equations by graphing?
3. x2-4x + 6 = 0
4. x2-4x-1=0
5. 4x2-12x + 3 = 0
6. x2-2x-4=0
You'll need a graphics calculator, and have the calculator plot those graphs.
The solutions are where the curve intersects the x-axis. You can refine the accuracy by zooming onto the intersection.
All of the above problems have two real solutions each, except for #3, which has no real zeroes (i.e. the roots are complex).
You are free to post your answers for a check if you wish. You can also check by solving them analytically using the quadratic equation.
If you do not have a graphics calculator, you can graph by paper and pencil with the same results.
To solve quadratic equations by graphing, you can follow these steps:
1. Rewrite the equation in the form of ax^2 + bx + c = 0, if it isn't already.
Now, let's solve the given quadratic equations by graphing:
3. x^2 - 4x + 6 = 0
Start by plotting the graph of the equation on a Cartesian plane. Since the coefficient of x^2 is positive (a = 1), the graph will be an upward-opening parabola.
Next, determine if the graph intersects the x-axis, meaning there are real solutions to the equation. If there are x-intercepts, those are the solutions to the equation.
If the graph does not intersect the x-axis, it means there are no real solutions.
4. x^2 - 4x - 1 = 0
Similarly, plot the graph of the equation. In this case, since a = 1, the parabola will also be upward-opening.
Check for x-intercepts to find the solutions.
5. 4x^2 - 12x + 3 = 0
Again, graph the equation. With a = 4, the parabola will be wider than the previous cases.
Look for x-intercepts to determine the solutions.
6. x^2 - 2x - 4 = 0
Plot the graph of the equation. This time, as a = 1, the parabola will be facing upwards.
Check for x-intercepts to find the solutions.
Remember, if a quadratic equation has no real solutions, the graph will not intersect the x-axis. In such cases, you can determine the nature of the solutions (real or complex) by using the discriminant (b^2 - 4ac).