Solving for 'a' in a Cubic Equation

The problem asks for the value of coefficient $a$ in the cubic equation $ax^3-24x^2+188x-480=0$. The roots are specified as three consecutive even natural numbers.

Using Vieta's Formulas

For a cubic equation $Ax^3+Bx^2+Cx+D=0$, Vieta's formulas relate the coefficients to the roots ($r_1, r_2, r_3$):

• Sum of roots: $r_1 + r_2 + r_3 = -B/A$
• Product of roots: $r_1r_2r_3 = -D/A$

In our equation, $A=a$, $B=-24$, $C=188$, $D=-480$. Thus:

• Sum: $r_1 + r_2 + r_3 = -(-24)/a = 24/a$
• Product: $r_1r_2r_3 = -(-480)/a = 480/a$

Determining the Roots

Let the three consecutive even natural numbers be $n-2$, $n$, and $n+2$. Since they are natural numbers, $n$ must be an even number $\ge 4$.

From the sum of the roots:

$ (n-2) + n + (n+2) = \frac{24}{a} $

$ 3n = \frac{24}{a} $

$ n = \frac{8}{a} $

Calculating the Value of 'a'

Using the product of the roots:

$ (n-2)(n)(n+2) = \frac{480}{a} $

$ n(n^2 - 4) = \frac{480}{a} $

Substitute $n = 8/a$ into the product equation:

$ \left(\frac{8}{a}\right) \left( \left(\frac{8}{a}\right)^2 - 4 \right) = \frac{480}{a} $

$ \frac{8}{a} \left( \frac{64}{a^2} - 4 \right) = \frac{480}{a} $

Assuming $a \ne 0$, multiply both sides by $a/8$:

$ \frac{64}{a^2} - 4 = \frac{480}{8} $

$ \frac{64}{a^2} - 4 = 60 $

$ \frac{64}{a^2} = 64 $

$ a^2 = 1 $

This implies $a = 1$ or $a = -1$.

Validating the Solution

Check $a=1$. If $a=1$, then $n = 8/1 = 8$. The roots are $n-2=6$, $n=8$, $n+2=10$. These (6, 8, 10) are consecutive even natural numbers.

Check $a=-1$. If $a=-1$, then $n = 8/(-1) = -8$. The roots would be $-10, -8, -6$. These are not natural numbers.

Therefore, the only valid value for $a$ is 1.