Question

Find the roots of the equation
2x^3-x^2-22x-24=0
if two of them are in the
ratio 4:3 .

Answer 1

Answer on Question #78503 – Math – Other

Question

Find the roots of the equation $2x^{3} - x^{2} - 22x - 24 = 0$ if two of them are in the ratio 4:3.

Solution

Let's denote roots as $x_{1}, x_{2} = 4a$ and $x_{3} = 3a$, where roots $x_{2}$ and $x_{3}$ satisfy the given relation. So we only have to find $x_{1}$ and $a$.

If the roots of the given equation are known then it can be written as follows

$$2 (x - x _ {1}) (x - x _ {2}) (x - x _ {3}) = 0$$

Substituting $x_{2} = 4a$ and $x_{3} = 3a$ and expanding we obtain

$$2 x ^ {3} - (2 x _ {1} + 1 4 a) x ^ {2} + (1 4 a x _ {1} + 2 4 a ^ {2}) x - 2 4 a ^ {2} x _ {1} = 0$$

Equating the coefficients for corresponding degrees of $x$ of this equation and the given one, we obtain

$$2 = 2$$ $$2 x _ {1} + 1 4 a = 1$$ $$1 4 a x _ {1} + 2 4 a ^ {2} = - 2 2$$ $$2 4 a ^ {2} x _ {1} = 2 4$$

From the second equation:

$$2 x _ {1} = 1 - 1 4 a$$

Substituting into the third equation we obtain

$$7 a (1 - 1 4 a) + 2 4 a ^ {2} = - 2 2$$ $$7 4 a ^ {2} - 7 a - 2 2 = 0$$ $$a = - \frac {1}{2}, \frac {2 2}{3 7}$$

Lets take $a = -1/2$. Since $a^2 x_1 = 1$ (4th equation), we obtain

$$x _ {1} = \frac {1}{a ^ {2}} = 4, x _ {2} = 4 a = - 2, x _ {3} = 3 a = - \frac {3}{2}$$

By substitution we can verify that these are indeed the roots of the given equation.

**Answer**: $4, -2, -3/2$.

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