Question

If ${\alpha }_1,{\alpha }_2,....,{\alpha }_6({\alpha }_1>0)$ 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

Answer: (C)

$160$

Given: equation $x^6-12x^5+a{x}^4-b{x}^3+c{x}^2-dx+64=0$, and ${\alpha }_1,{\alpha }_2,...,{\alpha }_6({\alpha }_1>0)$ are the roots.

To find: the value of $b$

Sol: equation $x^6-12x^5+a{x}^4-b{x}^3+c{x}^2-dx+64=0$ is an expanded form of $(a-b{)}^6$

And $(a-b{)}^6=a^6-6a^{5}c+15a^4{c}^2-20a^3{c}^3+15a^2{c}^4-6a{c}^5+c^6$

where $a=x,c^6=64⟹c=2$

Hence, $(x-2{)}^6=x^6-6(2)x^5+15(2{)}^2{x}^4-20(2{)}^3{x}^3+15(2{)}^4{x}^2-6(2{)}^{5}x+(2{)}^6⟹(x-2{)}^6=x^6-12x^5+60x^4-160x^3+240x^2-192x+64=0$

comparing this with given equation we find that the value of $b$ is $160$