If the roots of x^3 +3px^2 +3qx+r=0 are in harmonic progression, then prove that 2q^3 =r(3pq−r)


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 200353 by deleteduser12 last updated on 17/Nov/23

$If the roots of x^{3} + {3 px}^{2} + 3 qx + r = 0 are$ $in harmonic progression, then prove that$ ${2 q}^{3} = r (3 pq - r)$
	Answered by jabarsing last updated on 18/Nov/23
	$To prove that the equation $2q^3 = r(3pq - r)$ holds when the roots of $x^3 + 3px^2 + 3qx + r = 0$ are in harmonic progression, we can start by assuming that the roots are $a, \frac{1}{a},$ and $-a - \frac{1}{a}$, where $a \neq 0$. Using Vieta's formulas, we know that the sum of the roots is zero, so $a + \frac{1}{a} - a - \frac{1}{a} = 0$, which is true. Now, let's find the value of $q$ using the sum of the product of the roots taken two at a time. We have: $a \cdot \frac{1}{a} + a \cdot \left(-a - \frac{1}{a}\right) + \frac{1}{a} \cdot \left(-a - \frac{1}{a}\right) = 3q$ Simplifying this expression, we get: $1 - a^2 - 1 - \frac{1}{a^2} = 3q$ $-a^2 - \frac{1}{a^2} = 3q$ Multiplying both sides by $-1$, we have: $a^2 + \frac{1}{a^2} = -3q$ Now, let's find the value of $r$ using the product of the roots. We have: $a \cdot \frac{1}{a} \cdot \left(-a - \frac{1}{a}\right) = -r$ Simplifying this expression, we get: $-a - \frac{1}{a} = -r$ Multiplying both sides by $-1$, we have: $a + \frac{1}{a} = r$ Now, let's substitute the value of $r$ in terms of $a$ into the equation we found for $q$: $a^2 + \frac{1}{a^2} = -3q$ $2q = -\left(a^2 + \frac{1}{a^2}\right)$ Squaring both sides, we get: $4q^2 = \left(a^2 + \frac{1}{a^2}\right)^2$ Expanding the right side, we have: $4q^2 = a^4 + 2 + \frac{1}{a^4}$ Now, let's multiply both sides by $a^4$ to eliminate the fractions: $4a^4q^2 = a^8 + 2a^4 + 1$ Rearranging the terms, we get: $a^8 + 2a^4 - 4a^4q^2 + 1 = 0$ Now, notice that the equation $x^3 + 3px^2 + 3qx + r = 0$ has roots $a, \frac{1}{a},$ and $-a - \frac{1}{a}$. Therefore, we can rewrite the equation as: $(x - a)\left(x - \frac{1}{a}\right)\left(x + a + \frac{1}{a}\right) = 0$ Expanding this equation, we get: $x^3 + \left(a + \frac{1}{a}\right)x^2 + \left(-a - \frac{1}{a}\right)x - 1 = 0$ Comparing the coefficients of this equation with the original equation, we can see that: $a + \frac{1}{a} = -3p$ and $-a - \frac{1}{a} = r$ Substituting these values into the equation we obtained earlier, we have: $a^8 + 2a^4 - 4a^4q^2 + 1 = 0$ $(-3p)^4 + 2(-3p)^2 - 4(-3p)^2q^2 + 1 = 0$ Simplifying this equation, we get: $81p^4 + 18p^2 - 36p^2q^2 + 1 = 0$ Dividing both sides by $9$, we have: $9p^4 + 2p^2 - 4p^2q^2 + \frac{1}{9} = 0$ Now, notice that $9p^4$ and $\frac{1}{9}$ are perfect squares, so we can rewrite the equation as: \(\left$
	To prove that the equation $2 q^{3} = r (3 p q - r)$ holds when the roots of $x^{3} + 3 p x^{2} + 3 q x + r = 0$ are in harmonic progression, we can start by assuming that the roots are $a, \frac{1}{a},$ and $- a - \frac{1}{a}$ , where $a \neq 0$ . Using Vieta's formulas, we know that the sum of the roots is zero, so $a + \frac{1}{a} - a - \frac{1}{a} = 0$ , which is true. Now, let's find the value of $q$ using the sum of the product of the roots taken two at a time. We have: $a \cdot \frac{1}{a} + a \cdot (- a - \frac{1}{a}) + \frac{1}{a} \cdot (- a - \frac{1}{a}) = 3 q$ Simplifying this expression, we get: $1 - a^{2} - 1 - \frac{1}{a^{2}} = 3 q$ $- a^{2} - \frac{1}{a^{2}} = 3 q$ Multiplying both sides by $- 1$ , we have: $a^{2} + \frac{1}{a^{2}} = - 3 q$ Now, let's find the value of $r$ using the product of the roots. We have: $a \cdot \frac{1}{a} \cdot (- a - \frac{1}{a}) = - r$ Simplifying this expression, we get: $- a - \frac{1}{a} = - r$ Multiplying both sides by $- 1$ , we have: $a + \frac{1}{a} = r$ Now, let's substitute the value of $r$ in terms of $a$ into the equation we found for $q$ : $a^{2} + \frac{1}{a^{2}} = - 3 q$ $2 q = - (a^{2} + \frac{1}{a^{2}})$ Squaring both sides, we get: $4 q^{2} = {(a^{2} + \frac{1}{a^{2}})}^{2}$ Expanding the right side, we have: $4 q^{2} = a^{4} + 2 + \frac{1}{a^{4}}$ Now, let's multiply both sides by $a^{4}$ to eliminate the fractions: $4 a^{4} q^{2} = a^{8} + 2 a^{4} + 1$ Rearranging the terms, we get: $a^{8} + 2 a^{4} - 4 a^{4} q^{2} + 1 = 0$ Now, notice that the equation $x^{3} + 3 p x^{2} + 3 q x + r = 0$ has roots $a, \frac{1}{a},$ and $- a - \frac{1}{a}$ . Therefore, we can rewrite the equation as: $(x - a) (x - \frac{1}{a}) (x + a + \frac{1}{a}) = 0$ Expanding this equation, we get: $x^{3} + (a + \frac{1}{a}) x^{2} + (- a - \frac{1}{a}) x - 1 = 0$ Comparing the coefficients of this equation with the original equation, we can see that: $a + \frac{1}{a} = - 3 p$ and $- a - \frac{1}{a} = r$ Substituting these values into the equation we obtained earlier, we have: $a^{8} + 2 a^{4} - 4 a^{4} q^{2} + 1 = 0$ $(- 3 p)^{4} + 2 (- 3 p)^{2} - 4 (- 3 p)^{2} q^{2} + 1 = 0$ Simplifying this equation, we get: $81 p^{4} + 18 p^{2} - 36 p^{2} q^{2} + 1 = 0$ Dividing both sides by $9$ , we have: $9 p^{4} + 2 p^{2} - 4 p^{2} q^{2} + \frac{1}{9} = 0$ Now, notice that $9 p^{4}$ and $\frac{1}{9}$ are perfect squares, so we can rewrite the equation as: \(\left
	Commented by jabarsing last updated on 18/Nov/23
	$We can write the equation as follows: $(x - a)(x - \frac{1}{a})(x + a + \frac{1}{a}) = 0$ By multiplying this expression, we get: $(x^2 - ax - \frac{1}{a}x + \frac{1}{a^2})(x + a + \frac{1}{a}) = 0$ Simplifying this expression, we obtain: $x^3 + (a + \frac{1}{a})x^2 + (a^2 + 1 + \frac{1}{a^2})x + (a + \frac{1}{a}) = 0$ By comparing this expression with the original equation, we can expand the following relationships: $a + \frac{1}{a} = -3p$ $a^2 + 1 + \frac{1}{a^2} = -3q$ $a + \frac{1}{a} = -r$ Now, by adding the first and third equations, we get: $2(a + \frac{1}{a}) = -3p - r$ By substituting the value of $r$ from the fourth equation, we get: $2q = -3p - r$ By multiplying both sides of this equation by $2q$, we get: $4q^2 = -6pq - 2qr$ By adding this equation to the previous equation we obtained for $q$, we get: $4q^2 + 2q = -6pq - 2qr - 3pq + r(3pq - r)$ Simplifying this expression, we get: $2q(2q + 1) = r(3pq - r)$ And by dividing both sides of this equation by 2, we get: $q^2 + q = \frac{r(3pq - r)}{2}$ Finally, by multiplying both sides of this equation by 2, we get: $2q^3 = r(3pq - r)$ Thus, assuming that the roots of the equation are in a harmonic sequence, we have arrived at the equation $2q^3 = r(3pq - r)$.$
	We can write the equation as follows: $(x - a) (x - \frac{1}{a}) (x + a + \frac{1}{a}) = 0$ By multiplying this expression, we get: $(x^{2} - a x - \frac{1}{a} x + \frac{1}{a^{2}}) (x + a + \frac{1}{a}) = 0$ Simplifying this expression, we obtain: $x^{3} + (a + \frac{1}{a}) x^{2} + (a^{2} + 1 + \frac{1}{a^{2}}) x + (a + \frac{1}{a}) = 0$ By comparing this expression with the original equation, we can expand the following relationships: $a + \frac{1}{a} = - 3 p$ $a^{2} + 1 + \frac{1}{a^{2}} = - 3 q$ $a + \frac{1}{a} = - r$ Now, by adding the first and third equations, we get: $2 (a + \frac{1}{a}) = - 3 p - r$ By substituting the value of $r$ from the fourth equation, we get: $2 q = - 3 p - r$ By multiplying both sides of this equation by $2 q$ , we get: $4 q^{2} = - 6 p q - 2 q r$ By adding this equation to the previous equation we obtained for $q$ , we get: $4 q^{2} + 2 q = - 6 p q - 2 q r - 3 p q + r (3 p q - r)$ Simplifying this expression, we get: $2 q (2 q + 1) = r (3 p q - r)$ And by dividing both sides of this equation by 2, we get: $q^{2} + q = \frac{r (3 p q - r)}{2}$ Finally, by multiplying both sides of this equation by 2, we get: $2 q^{3} = r (3 p q - r)$ Thus, assuming that the roots of the equation are in a harmonic sequence, we have arrived at the equation $2 q^{3} = r (3 p q - r)$ .
	Answered by ajfour last updated on 17/Nov/23

	$(x - \frac{1}{h}) (x - \frac{2}{h + k}) (x - \frac{1}{k}) = 0$ $\frac{2}{(h + k)} (\frac{1}{h} + \frac{1}{k}) + \frac{1}{h k} = 3 q$ $\Rightarrow h k = \frac{1}{q} . . . (i)$ $\frac{1}{h} + \frac{1}{k} + \frac{2}{h + k} = - 3 p$ $\Rightarrow q {(h + k)}^{2} + 2 + 3 p (h + k) = 0 . . (i i)$ $& \frac{1}{h k} (\frac{2}{h + k}) = - r$ $\Rightarrow h + k = - \frac{2 q}{r} . . . (i i i)$ $U s i n g (i i i) i n (i i)$ $q {(- \frac{2 q}{r})}^{2} + 2 = 3 p (\frac{2 q}{r})$ $2 q^{3} + r^{2} = 3 p q r$ $\Rightarrow 2 q^{3} = r (3 p q - r)$
	Answered by Rasheed.Sindhi last updated on 17/Nov/23

	$l e t a, \frac{2 a b}{a + b}, b a r e t h e r o o t s$ $∙ a + \frac{2 a b}{a + b} + b = - 3 p . . . . . . . . . (A)$ $∙ a (\frac{2 a b}{a + b}) + (\frac{2 a b}{a + b}) (b) + b a = 3 q$ $\frac{2 a^{2} b + 2 {a b}^{2}}{a + b} + b a = 3 q$ $\frac{2 a b (a + b)}{a + b} = 3 q$ $3 a b = 3 q \Rightarrow a b = q \begin{matrix} a b = q \end{matrix}$ $∙ a (\frac{2 a b}{a + b}) (b) = - r$ $\frac{2 a^{2} b^{2}}{a + b} = - r \Rightarrow \frac{2 {(a b)}^{2}}{a + b} = - r$ $a + b = \frac{2 q^{2}}{- r} = - \frac{2 q^{2}}{r}$ $\begin{matrix} a + b = - \frac{2 q^{2}}{r} \end{matrix}$ $(A) : a + b + \frac{2 a b}{a + b} = - 3 p$ $- \frac{2 q^{2}}{r} + \frac{2 q}{- \frac{2 q^{2}}{r}} = - 3 p$ $- \frac{2 q^{2}}{r} - \frac{2 q r}{2 q^{2}} = - 3 p$ $\frac{2 q^{2}}{r} + \frac{r}{q} = 3 p$ $3 p q r = 2 q^{3} + r^{2}$ $2 q^{3} = r (3 p q - r)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum