Question

If the set of all values of \( a \), for which the equation \( 5x^3 - 15x - a = 0 \) has three distinct real roots, is the interval \( (\alpha, \beta) \), then \( \beta - 2\alpha \) is equal to                

Hint:

To determine the conditions for three distinct real roots, find the critical points of the function and use the graph to determine the appropriate range of values for \( a \).

Answer 1

The given equation is: \[ 5x^3 - 15x - a = 0 \] Let \( f(x) = 5x^3 - 15x \). 
Now, differentiate \( f(x) \): \[ f'(x) = 15x^2 - 15 = 15(x - 1)(x + 1) \] Thus, the critical points of the function are \( x = 1 \) and \( x = -1 \). 
Next, to find the condition for three distinct real roots, we need to find the values of \( a \) such that the graph of \( f(x) \) intersects the x-axis at three points. 
Plotting the function shows that the values of \( a \) must lie in the interval \( (-10, 10) \). 
Therefore, \( \alpha = -10 \) and \( \beta = 10 \). 
Finally, calculate \( \beta - 2\alpha \): \[ \beta - 2\alpha = 10 - 2(-10) = 10 + 20 = 30 \] Thus, the value of \( \beta - 2\alpha \) is \( 30 \).

Answer 2

Given $f(x)=5x^3-15x$. Then \[ f'(x)=15x^2-15=15(x-1)(x+1), \] so critical points are $x=\pm1$. 

Evaluate $f$ at those points: \[ f(1)=5(1)^3-15(1)=-10,\qquad f(-1)=5(-1)^3-15(-1)=10. \] For the cubic $5x^3-15x-a=0$ to have three distinct real roots, the horizontal line $y=a$ must cut the cubic in three places, so $a$ must lie strictly between $f(1)$ and $f(-1)$: \[ a\in(-10,\,10). \] Thus $\alpha=-10$ and $\beta=10$.

Required value

\[ \beta-2\alpha=10-2(-10)=10+20=30. \]

Answer:

$\boxed{30}$