Question

if the roots of equation 10x3-cx2-54x-27=0 are in hp find c and roots if the roots of equation 10x3-cx2-54x-27=0 are in hp find c and roots

Answer 1

c=9, roots are 3, -3/2, -3/5

Answer 2

Given the equation: $10x^3 - cx^2 - 54x - 27 = 0$. Let the roots of this equation be $\alpha, \beta, \gamma$. Since the roots are in Harmonic Progression (HP), their reciprocals, $1/\alpha, 1/\beta, 1/\gamma$, will be in Arithmetic Progression (AP).

To find an equation whose roots are $1/\alpha, 1/\beta, 1/\gamma$, we substitute $x = 1/y$ into the original equation: $10(1/y)^3 - c(1/y)^2 - 54(1/y) - 27 = 0$ $\frac{10}{y^3} - \frac{c}{y^2} - \frac{54}{y} - 27 = 0$ Multiply the entire equation by $y^3$ to clear denominators: $10 - cy - 54y^2 - 27y^3 = 0$ Rearrange in standard cubic form: $-27y^3 - 54y^2 - cy + 10 = 0$ Or, $27y^3 + 54y^2 + cy - 10 = 0$

Let the roots of this new equation be $y_1, y_2, y_3$. These roots are $1/\alpha, 1/\beta, 1/\gamma$ and are in AP. Let the roots in AP be $A-D, A, A+D$. According to Vieta's formulas, the sum of the roots of $27y^3 + 54y^2 + cy - 10 = 0$ is: $y_1 + y_2 + y_3 = (A-D) + A + (A+D) = -\frac{54}{27}$ $3A = -2$ $A = -\frac{2}{3}$

So, the middle root of the AP is $y_2 = A = -2/3$. Since $y_2 = 1/\beta$, we have $1/\beta = -2/3$, which means $\beta = -3/2$. This is one of the roots of the original equation $10x^3 - cx^2 - 54x - 27 = 0$.

Substitute $x = -3/2$ into the original equation to find the value of $c$: $10(-3/2)^3 - c(-3/2)^2 - 54(-3/2) - 27 = 0$ $10(-27/8) - c(9/4) + 81 - 27 = 0$ $-\frac{270}{8} - \frac{9c}{4} + 54 = 0$ $-\frac{135}{4} - \frac{9c}{4} + \frac{216}{4} = 0$ Multiply by 4: $-135 - 9c + 216 = 0$ $81 - 9c = 0$ $9c = 81$ $c = 9$

Now the equation is $10x^3 - 9x^2 - 54x - 27 = 0$. We know one root is $x = -3/2$. We can use polynomial division or synthetic division to find the other roots. Using synthetic division with root $-3/2$:

-3/2 | 10   -9   -54   -27
     |      -15    36    27
     --------------------
       10   -24   -18     0

The quotient is $10x^2 - 24x - 18 = 0$. Divide by 2: $5x^2 - 12x - 9 = 0$. Now solve this quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(5)(-9)}}{2(5)}$ $x = \frac{12 \pm \sqrt{144 + 180}}{10}$ $x = \frac{12 \pm \sqrt{324}}{10}$ $x = \frac{12 \pm 18}{10}$

The two remaining roots are: $x_1 = \frac{12 + 18}{10} = \frac{30}{10} = 3$ $x_2 = \frac{12 - 18}{10} = \frac{-6}{10} = -3/5$

So, the roots of the equation are $3, -3/2, -3/5$. To verify they are in HP, check their reciprocals: $1/3, -2/3, -5/3$. These values form an AP with a common difference of $-1$. $(-2/3) - (1/3) = -3/3 = -1$ $(-5/3) - (-2/3) = -3/3 = -1$ Thus, the roots are indeed in HP.

The value of $c$ is 9. The roots are $3, -3/2, -3/5$.