Question

If $\alpha ={\log }_{3}4$ is one of the roots of equation $x^3+a{x}^2+bx+1=0$, $a,b\in R$, then equation which always have a root $\beta ={\log }_{2}48$, is

• A. $x^3+2(b-6)x^2+4(a-4b+12)x+8(4b-2a-7)=0$
• B. $x^3+b{x}^2+ax+1=0$
• C. $(65+16a+4b)x^2+2(48+8a+b)x^2+4(12+a)x+8=0$
• D. $(65+16a+4b)x^2+(48+8a+b)x^2+(12+a)x+8=0$

Answer 1

Answer: (A)

$x^3+2(b-6)x^2+4(a-4b+12)x+8(4b-2a-7)=0$

$\alpha ={\log }_{3}4=\frac{1}{{\log }_{4}3}=\frac{1}{\frac{1}{2}{\log }_{2}3}=\frac{2}{{\log }_{2}3}$

$\beta ={\log }_{2}48={\log }_2(2^4\times 3)=4+{\log }_{2}3=4+\frac{2}{\alpha }$

$⟹\frac{2}{\alpha }=\beta -4$

$⟹\alpha =\frac{2}{\beta -4}$

Since $\alpha$ is a root of equation $x^3+a{x}^2+bx+1=0$

$⟹{\alpha }^3+a{\alpha }^2+b\alpha +1=0$

$⟹\frac{8}{(\beta -4{)}^3}+\frac{4a}{(\beta -4{)}^2}+\frac{2b}{\beta -4}+1=0$

$⟹8+4a(\beta -4)+2b(\beta -4{)}^2+(\beta -4{)}^3=0$

$⟹8+4a(\beta -4)+2b({\beta }^2-8\beta +16)+({\beta }^3-12{\beta }^2+48\beta -64)=0$

$⟹{\beta }^3+{\beta }^2(2b-12)+\beta (4a-16b+48)+(32b-16a-56)=0$

$⟹{\beta }^3+2{\beta }^2(b-6)+4\beta (a-4b+12)+8(4b-2a-7)=0$

putting $\beta =x$, we get-

$x^3+2(b-6)x^2+4(a-4b+12)x+8(4b-2a-7)=0$

Hence,answer is (A).