What is the solution to the linear quadratic system of equations given by y = 25x^3 and y = x^2?

by

To solve the system of equations where y = 25x3 and y = x2, we first need to set the two expressions for y equal to each other:

25x3 = x2

Next, we can rearrange this equation by moving all terms to one side:

25x3 - x2 = 0

Now, we can factor out the common term:

x2(25x - 1) = 0

This equation will be true if either factor equals zero. Therefore, we can solve for x:

  1. x2 = 0:
    • This gives us x = 0.
  2. 25x – 1 = 0:
    • Solving for x, we get:
      • x = \frac{1}{25}

Now that we have the values for x, we can find the corresponding values of y by substituting these x values into either of the original equations. Let’s start with x = 0:

y = 25(0)3 = 0

So, one solution is:

(0, 0)

Now, substituting x = \frac{1}{25}: 

y = 25(\frac{1}{25})3 = 25(\frac{1}{15625}) = \frac{1}{625}

This gives us another solution:

(\frac{1}{25}, \frac{1}{625})

In summary, the solutions to the given system of equations are:

  • (0, 0)
  • (\frac{1}{25}, \frac{1}{625})

These points represent the intersections of the curves defined by the two equations.

Leave a Comment