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

applicable. Use the solution method of your choice.

a) 4𝑥^2 = 4𝑥 + 15

1 answer

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

4𝑥^2 - 4𝑥 - 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 𝑎𝑥^2 + 𝑏𝑥 + 𝑐 = 0, the solutions are given by:

𝑥 = (-𝑏 ± √(𝑏^2 - 4𝑎𝑐)) / (2𝑎)

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

𝑥 = (-(-4) ± √((-4)^2 - 4(4)(-15))) / (2(4))
𝑥 = (4 ± √(16 + 240)) / 8
𝑥 = (4 ± √256) / 8
𝑥 = (4 ± 16) / 8

Taking the positive and negative values, we have:

𝑥₁ = (4 + 16) / 8 = 20 / 8 = 2.5
𝑥₂ = (4 - 16) / 8 = -12 / 8 = -1.5

So the solutions to the equation 4𝑥^2 = 4𝑥 + 15 are 𝑥 = 2.5 and 𝑥 = -1.5.