Question

If ${\alpha }_1,{\alpha }_2......{\alpha }_6({\alpha }_i>0)$ are the roots of the equation $x^6-12x^5+a{x}^4-b{x}^3+c{x}^2-dx+64=0$ then b is equal to

• A. $20$
• B. $60$
• C. $160$
• D. $240$

Answer 1

The correct option is C $160$

The given equation is sixth order equation. Thus, given equation will have six roots.

Let ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5,{\alpha }_6$ are positive roots of the given equation.

We know that, Sum of all roots of equation = $\frac{-b}{a}$

Where, b= coefficient of $x^5$ and a= coefficient of $x^6$

$\therefore ({\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4+{\alpha }_5+{\alpha }_6)=\frac{-(-12)}{1}=12$

Similarly, Product of all roots of equation = $\frac{e}{a}$

Where, e= Constant term in the equation

$\therefore \left({\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4{\alpha }_5{\alpha }_6\right)=\frac{64}{1}=64$

Now, we know that Arithmetic Mean of roots $\ge$ Geometric Mean of roots

Arithmetic Mean of roots = $\frac{{\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4+{\alpha }_5+{\alpha }_6}{6}=\frac{12}{6}=2$

Geometric Mean of roots = ${({\alpha }_1.{\alpha }_2.{\alpha }_3.{\alpha }_4.{\alpha }_5.{\alpha }_6)}^{\frac{1}{6}}={\left(64\right)}^{\frac{1}{6}}=2$

Thus, A.M. = G. M.

Thus, all the roots of the equation are equal and value of each root is 2.

$\therefore x^6-12x^5+a{x}^4-b{x}^3+c{x}^2-dx+64={(x-2)}^6$

$\therefore x^6-12x^5+a{x}^4-b{x}^3+c{x}^2-dx+64={\left({(x-2)}^2\right)}^3$

$\therefore x^6-12x^5+a{x}^4-b{x}^3+c{x}^2-dx+64={(x^2-4x+4)}^3$

$\therefore x^6-12x^5+a{x}^4-b{x}^3+c{x}^2-dx+64={\left(x\right(x-4)+4)}^3$

We know that ${(a+b)}^3=a^3+3a^{2}b+3a{b}^2+b^3$

Thus, above equation becomes,

$x^6-12x^5+a{x}^4-b{x}^3+c{x}^2-dx+64={\left[x\right(x-4\left)\right]}^3+[3{\times \left(x\right(x-4\left)\right)}^2\times 4]+[3\times x(x-4)\times {\left(4\right)}^2]+{\left(4\right)}^3$

$\therefore x^6-12x^5+a{x}^4-b{x}^3+c{x}^2-dx+64=x^3{(x-4)}^3+12x^2{(x-4)}^2+48x(x-4)+64$

$\therefore x^6-12x^5+a{x}^4-b{x}^3+c{x}^2-dx+64=x^3(x^3-12x^2+48x-64)+12x^2(x^2-8x+16)+48(x^2-4x)+64$

$\therefore x^6-12x^5+a{x}^4-b{x}^3+c{x}^2-dx+64=x^6-12x^5+48x^4-64x^3+12x^4-96x^3+192x^2+48x^2-192x+64$

$\therefore x^6-12x^5+a{x}^4-b{x}^3+c{x}^2-dx+64=x^6-12x^5+60x^4-160x^3+240x^2-192x+64$

Comparing coefficients of $x^6,x^5,x^4,x^3,x^2,x$ and constant terms on LHS and RHS, we get,

$a=60$, $b=160$, $c=240$, $d=192$

Thus, answer is option (C)