Question

11.
If the roots of x3 + 3ax? + 3bx +C =0 are in H.P. then
1) 2b^2 = c(3ab -c)
2) 2b^3= c(3ab -c)
3) 2b^3 = c² (3ab -c) 4) 2b = c2 (3ab -c)​

Answer 1

Correct-Question

If the roots of $x^{3} +3ax^{2} +3bx+c=0$ are in H.P. then

1) $2b^2 = c(3ab -c)$

2) $2b^3= c(3ab -c)$

3) $2b^3 = c^{2} (3ab -c) 4) 2b = c2 (3ab -c)$

Answer:

If the roots of$x^{3} +3ax^{2} +3bx+c=0$ are in H.P. then $2b^{3} =c(3ab-c)$ and option (2) is correct.

Step-by-step explanation:

Given:

The roots of polynomial  $x^{3} +3ax^{2} +3bx+c=0$ are in H.P.

To Find:

If the roots of $x^{3} +3ax^{2} +3bx+c=0$ are in H.P. then find out relation between a,b and c.

Solution:

As given-the roots of polynomial  $x^{3} +3ax^{2} +3bx+c=0$ are in H.P.

Let roots of polynomial are r,s  and t.

r.s and t are in H.P.

Therefore,  $s=\frac{2rt}{r+t}$

$sr+st=2rt$

adding rt on both side of equation.

$sr+st+rt=3rt$  ------------- equation no.01.

Using the theory of polynomials, in polynomial $x^{3} +3ax^{2} +3bx+c=0$.

$rst=\frac{-c}{1} \\rst=-c$ --------equation no.02.

$rs+st+rt=\frac{3b}{1} \\\\rs+st+rt =3b$  ----------- equation no.03.

Putting value of $rs+st+tr$ from equation no.01 to equation no.03. We get.

$3rt= 3b$

$rt=b$ -------------- equation no.04

Putting the value of b from equation no.04  to equation no.02. We get.

$b\times s= - c$

$s=\frac{-c}{b}$

Since  s is a root of the equation $x^{3} +3ax^{2} +3bx+c=0$.

ence,we can substitute the value of s in the equation.

$(\frac{-c}{b} )^{3} +3a(\frac{-c}{b}})^{2} +3bc+c=0$

$\frac{-c^{3} }{b^{3} } +\frac{3ac^{2} }{b^{2} } -\frac{3bc}{b}+c=0}$

$\frac{-c^{3}+3abc^{2}-3b^{3}c+cb^{3} }{b^{3} } =0$

$-c^{3}+3abc^{2}-3b^{3}c+cb^{3} =0$

$-c^{3}+3abc^{2}-2b^{3}c =0$

Dividing by c on the both sides

$-c^{2}+3abc-2b^{3} =0$

$2b^{3} =3abc-c^{2}$

$2b^{3} = c(3ab-c)$

Thus,If the roots of$x^{3} +3ax^{2} +3bx+c=0$ are in H.P. then $2b^{3} =c(3ab-c)$ and option (2) is correct.

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Answer 2

Answer:

If the roots of $x^{3} + 3a x^{2} + 3bx +c =0$ are in H.P, then $2b^{3}=c(3ab -c)$ .

Step-by-step explanation:

Let the roots of the equation be $\alpha, \beta$ and $\gamma$. Since they are in H.P,

$\therefore$  $\beta=\frac{2\alpha\gamma}{\alpha+\gamma}$

$\implies$ $\alpha \beta +\beta \gamma=2\alpha \gamma$

$\implies \alpha \beta +\beta \gamma+\alpha \gamma=3\alpha \gamma$      .....(i)

We know, the product of roots : $\alpha \beta \gamma=-c$

$\implies\alpha \gamma=\frac{-c}{\beta }$        ....(ii)

Putting the value of $\alpha\gamma$ from eq (ii) and eq (i)

$\implies \alpha \beta +\beta \gamma+\alpha \gamma=3\frac{-c}{\beta }$       ....(iii)

Also, sum of products :  $\alpha \beta +\beta \gamma+\alpha \gamma=3b$    ...(iv)

Comparing eq (iii) and eq (iv)

$3b=-3\frac{c}{\beta }$          

 $\implies \beta =\frac{-c}{b}$         .........(v)

Putting the value of $\beta$  from eq (v) in the given equation $x^{3} + 3a x^{2} + 3bx +c =0$, we get

$\frac{-c^{3} }{b^{3} } +\frac{3ac^{2} }{b^{2} } -3c+c=0\\\\\implies -c^{3}+3abc^{2}=2cb^{3}\\\\\\\implies 2b^{3}=c(3ab -c)$

Hence, if the roots of  $x^{3} + 3a x^{2} + 3bx +c =0$ are in H.P, then $2b^{3}=c(3ab -c)$ .

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