If-the-roots-of-x-3-3px-2-3qx-r-0-are-in-harmonic-progression-then-prove-that-2q-3-r-3pq-r-

Question Number 200353 by faysal last updated on 17/Nov/23

$$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} +\mathrm{3px}^{\mathrm{2}} +\mathrm{3qx}+\mathrm{r}=\mathrm{0}\:\mathrm{are} \\ $$$$\mathrm{in}\:\mathrm{harmonic}\:\mathrm{progression},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{2q}^{\mathrm{3}} =\mathrm{r}\left(\mathrm{3pq}−\mathrm{r}\right) \\ $$

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:

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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 $2q^3 = r(3pq – r)$.

Answered by ajfour last updated on 17/Nov/23

(x−(1/h))(x−(2/(h+k)))(x−(1/k))=0 (2/((h+k)))((1/h)+(1/k))+(1/(hk))=3q ⇒ hk=(1/q) ...(i) (1/h)+(1/k)+(2/(h+k))=−3p ⇒ q(h+k)^2 +2+3p(h+k)=0 ..(ii) & (1/(hk))((2/(h+k)))=−r ⇒ h+k=−((2q)/r) ...(iii) Using (iii) in (ii) q(−((2q)/r))^2 +2=3p(((2q)/r)) 2q^3 +r^2 =3pqr ⇒ 2q^3 =r(3pq−r)

$$\left({x}−\frac{\mathrm{1}}{{h}}\right)\left({x}−\frac{\mathrm{2}}{{h}+{k}}\right)\left({x}−\frac{\mathrm{1}}{{k}}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{2}}{\left({h}+{k}\right)}\left(\frac{\mathrm{1}}{{h}}+\frac{\mathrm{1}}{{k}}\right)+\frac{\mathrm{1}}{{hk}}=\mathrm{3}{q} \\ $$$$\Rightarrow\:\:{hk}=\frac{\mathrm{1}}{{q}}\:\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{1}}{{h}}+\frac{\mathrm{1}}{{k}}+\frac{\mathrm{2}}{{h}+{k}}=−\mathrm{3}{p} \\ $$$$\Rightarrow\:\:{q}\left({h}+{k}\right)^{\mathrm{2}} +\mathrm{2}+\mathrm{3}{p}\left({h}+{k}\right)=\mathrm{0}\:\:..\left({ii}\right) \\ $$$$\&\:\:\:\frac{\mathrm{1}}{{hk}}\left(\frac{\mathrm{2}}{{h}+{k}}\right)=−{r} \\ $$$$\Rightarrow\:\:{h}+{k}=−\frac{\mathrm{2}{q}}{{r}}\:\:…\left({iii}\right) \\ $$$${Using}\:\left({iii}\right)\:\:{in}\:\:\left({ii}\right) \\ $$$${q}\left(−\frac{\mathrm{2}{q}}{{r}}\right)^{\mathrm{2}} +\mathrm{2}=\mathrm{3}{p}\left(\frac{\mathrm{2}{q}}{{r}}\right) \\ $$$$\mathrm{2}{q}^{\mathrm{3}} +{r}^{\mathrm{2}} =\mathrm{3}{pqr} \\ $$$$\Rightarrow\:\:\:\mathrm{2}{q}^{\mathrm{3}} ={r}\left(\mathrm{3}{pq}−{r}\right) \\ $$

Answered by Rasheed.Sindhi last updated on 17/Nov/23

let a , ((2ab)/(a+b)) , b are the roots •a+((2ab)/(a+b))+b=−3p.........(A) •a(((2ab)/(a+b)))+(((2ab)/(a+b)))(b)+ba=3q ((2a^2 b+2ab^2 )/(a+b))+ba=3q ((2ab(a+b))/(a+b))=3q 3ab=3q⇒ab=q determinant (((ab=q))) •a(((2ab)/(a+b)))(b)=−r ((2a^2 b^2 )/(a+b))=−r⇒((2(ab)^2 )/(a+b))=−r a+b=((2q^2 )/(−r))=−((2q^2 )/r) determinant (((a+b=−((2q^2 )/r)))) (A): a+b+((2ab)/(a+b))=−3p −((2q^2 )/r)+((2q)/(−((2q^2 )/r)))=−3p −((2q^2 )/r)−((2qr)/(2q^2 ))=−3p ((2q^2 )/r)+(r/q)=3p 3pqr=2q^3 +r^2 2q^3 =r(3pq−r)

$${let}\:{a}\:,\:\frac{\mathrm{2}{ab}}{{a}+{b}}\:,\:{b}\:{are}\:{the}\:{roots} \\ $$$$\bullet{a}+\frac{\mathrm{2}{ab}}{{a}+{b}}+{b}=−\mathrm{3}{p}………\left({A}\right) \\ $$$$\bullet{a}\left(\frac{\mathrm{2}{ab}}{{a}+{b}}\right)+\left(\frac{\mathrm{2}{ab}}{{a}+{b}}\right)\left({b}\right)+{ba}=\mathrm{3}{q} \\ $$$$\:\:\:\:\frac{\mathrm{2}{a}^{\mathrm{2}} {b}+\mathrm{2}{ab}^{\mathrm{2}} }{{a}+{b}}+{ba}=\mathrm{3}{q} \\ $$$$\:\:\:\frac{\mathrm{2}{ab}\left({a}+{b}\right)}{{a}+{b}}=\mathrm{3}{q} \\ $$$$\:\:\:\mathrm{3}{ab}=\mathrm{3}{q}\Rightarrow{ab}={q}\:\begin{array}{|c|}{{ab}={q}}\\\hline\end{array} \\ $$$$\bullet{a}\left(\frac{\mathrm{2}{ab}}{{a}+{b}}\right)\left({b}\right)=−{r} \\ $$$$\:\:\:\frac{\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}+{b}}=−{r}\Rightarrow\frac{\mathrm{2}\left({ab}\right)^{\mathrm{2}} }{{a}+{b}}=−{r} \\ $$$${a}+{b}=\frac{\mathrm{2}{q}^{\mathrm{2}} }{−{r}}=−\frac{\mathrm{2}{q}^{\mathrm{2}} }{{r}} \\ $$$$\begin{array}{|c|}{{a}+{b}=−\frac{\mathrm{2}{q}^{\mathrm{2}} }{{r}}}\\\hline\end{array} \\ $$$$\left({A}\right):\:\:{a}+{b}+\frac{\mathrm{2}{ab}}{{a}+{b}}=−\mathrm{3}{p} \\ $$$$−\frac{\mathrm{2}{q}^{\mathrm{2}} }{{r}}+\frac{\mathrm{2}{q}}{−\frac{\mathrm{2}{q}^{\mathrm{2}} }{{r}}}=−\mathrm{3}{p} \\ $$$$−\frac{\mathrm{2}{q}^{\mathrm{2}} }{{r}}−\frac{\mathrm{2}{qr}}{\mathrm{2}{q}^{\mathrm{2}} }=−\mathrm{3}{p} \\ $$$$\frac{\mathrm{2}{q}^{\mathrm{2}} }{{r}}+\frac{{r}}{{q}}=\mathrm{3}{p} \\ $$$$\mathrm{3}{pqr}=\mathrm{2}{q}^{\mathrm{3}} +{r}^{\mathrm{2}} \\ $$$$\mathrm{2}{q}^{\mathrm{3}} ={r}\left(\mathrm{3}{pq}−{r}\right) \\ $$

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