Question

If $1,2,3$ are the roots of $a{x}^3+b{x}^2+cx+d=0$, then the roots of $ax\sqrt{x}+bx+c\sqrt{x}+d=0$, are

• A. $2,3,4$
• B. $1,4,9$
• C. $2,4,6$
• D. $\sqrt{2},\sqrt{3},\sqrt{5}$

Answer 1

Answer: (B)

$1,4,9$

As $1,2,3$ are roots of $a{x}^3+b{x}^2+cx+d=0$

Then $S_1=1+2+3=6=-\frac{b}{a}$

$S_2=1\times 2+1\times 3+2\times 3=11=\frac{c}{a}$

$S_3=1\times 2\times 3=6=-\frac{d}{a}$

Now $ax\sqrt{x}+bx+c\sqrt{x}+d=0\Rightarrow x\sqrt{x}+\frac{b}{a}x+\frac{c}{a}\sqrt{x}+\frac{d}{a}=0$

$\Rightarrow x\sqrt{x}-6x+11\sqrt{x}-6=0\Rightarrow x\sqrt{x}+11\sqrt{x}=6x+6\Rightarrow {(x\sqrt{x}+11\sqrt{x})}^2={(6x+6)}^2\Rightarrow x^3+121x+22x^2=36x^2+36+72x\Rightarrow x^3-14x^2+49x-36=0\Rightarrow (x-1)(x-4)(x-9)=0$

Hence, the roots are $1,4,9$.