Determine the number of real-number solutions

Determine the number of real-number solutions to the equation from the given graph 4x2 + 1 = 4x, given the graph of y = 4x -4x2 -1.
STEP 1:   We note that both the given equations are in the variable x, and are of the quadratic form ax2 + bx + c. Recall the fact that the solutions of an equation of the form ax2 + bx +c = 0 are its x-intercepts.

The solutions of such equations can be easily found out by sketching the respective graphs. To be clearer we graph each of the given equation, and then identify the points where the graph intersects the x-axis.

We shall first sketch the graph of 4x2 + 1 = 4x. Here, the point to be noticed is that to plot the equation we rewrite it in the form
y = 4x2 – 4x + 1.
quadratic curve y=4x^2-4x+1

STEP 2:    We have already discussed the fact that the solution of the equation is the x-value at which the respective graph intersects the x-axis. From the graph, it is clear that it intersects the x-axis at

STEP 3:    The equation we have to graph now is y = 4x2 – 4x + 1. We plot the graph similar to what we have done before.
quadratic curve y=4x-4x^2-1

Again, it is found that the graph intersects the x-axis at the same point x = 0.5. So, here also the solution to the equation is x = 0.5.

There is only one real number solution to both the equations.

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