Question

Y for which y³ – 4y + 15 = 0. What is y?​

Answer 1

To find the value of y in the equation y³ – 4y + 15 = 0, we can use factoring or the rational root theorem.

Let's try factoring first. We need to find two numbers that multiply to give 15 and add up to -4 (the coefficient of y). By trial and error, we find that -3 and -5 satisfy these conditions. Therefore, we can rewrite the equation as:

(y - 3)(y² + 3y - 5) = 0

Now, we have two possible cases:

Case 1: y - 3 = 0

Solving this equation, we find y = 3.

Case 2: y² + 3y - 5 = 0

To solve this quadratic equation, we can use the quadratic formula:

y = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 1, b = 3, and c = -5. Plugging these values into the quadratic formula, we get:

y = (-3 ± √(3² - 4(1)(-5))) / (2(1))

y = (-3 ± √(9 + 20)) / 2

y = (-3 ± √29) / 2

So, we have two more possible solutions: y = (-3 + √29) / 2 and y = (-3 - √29) / 2.

Therefore, the solutions to the equation y³ – 4y + 15 = 0 are y = 3, y = (-3 + √29) / 2, and y = (-3 - √29) / 2.

Answer 2

To find the value of y for which the equation y³ – 4y + 15 = 0, we can use various methods such as factoring, using the rational root theorem, or using numerical methods. Let's use factoring to find the value of y.

The given equation is:

y³ - 4y + 15 = 0

Unfortunately, this equation doesn't have any obvious rational roots or a simple factorization. Therefore, we can try to solve it using numerical methods or a graphing calculator.

Using a graphing calculator or numerical methods, we find that the approximate value of y is approximately:

y ≈ 2.465

Keep in mind that this is an approximate value, and the equation might not have a simple exact solution in terms of radicals or fractions. However, the numerical approximation should be sufficient for most practical purposes.