Solve each of the following equations. State the exact and approximate answers as

applicable. Use the solution method of your choice.

a) 4π‘₯^2 = 4π‘₯ + 15

To solve the equation 4π‘₯^2 = 4π‘₯ + 15, we can rearrange it to form a quadratic equation:

4π‘₯^2 - 4π‘₯ - 15 = 0

We can solve this equation by factoring or by using the quadratic formula. Let's use the quadratic formula:

The quadratic formula states that for an equation of the form π‘Žπ‘₯^2 + 𝑏π‘₯ + 𝑐 = 0, the solutions are given by:

π‘₯ = (-𝑏 Β± √(𝑏^2 - 4π‘Žπ‘)) / (2π‘Ž)

In our equation, π‘Ž = 4, 𝑏 = -4, and 𝑐 = -15. Plugging these values into the quadratic formula, we get:

π‘₯ = (-(-4) Β± √((-4)^2 - 4(4)(-15))) / (2(4))
π‘₯ = (4 Β± √(16 + 240)) / 8
π‘₯ = (4 Β± √256) / 8
π‘₯ = (4 Β± 16) / 8

Taking the positive and negative values, we have:

π‘₯₁ = (4 + 16) / 8 = 20 / 8 = 2.5
π‘₯β‚‚ = (4 - 16) / 8 = -12 / 8 = -1.5

So the solutions to the equation 4π‘₯^2 = 4π‘₯ + 15 are π‘₯ = 2.5 and π‘₯ = -1.5.