Question

The roots of the equation $x^4-8x^3+a{x}^2-bx+16=0$, are positive, if

• A. $a=24$
• B. $a=12$
• C. $b=8$
• D. $b=32$

Answer 1

The correct option is A $a=24$

$x^4-8x^3+a{x}^2-bx+16=0$

${\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4=8$

${\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4=16$

$AM=GM=2$

Dividing from $AM-GM$ inequality

$(\frac{{\sum }_1^{n}i=1}{n}\ge {\left({\prod }_{i=1}^{n}ai\right)}^{1/n})$

Only when all the terms are equal .

Hence all roots are equal

So ${\alpha }_1={\alpha }_2={\alpha }_3={\alpha }_4=2$

So, $A$ and $B$ are true