Question

If 1, 5, 7 are the roots of \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\], then the roots of \[a{{x}^{3}}+2b{{x}^{2}}+4cx+8d=0\] are:
(a) 2, 10, 14
(b) $\dfrac{1}{2}$, $\dfrac{5}{2}$, $\dfrac{7}{2}$
(c) –2, –10, –14
(d) 0, 2, 10

Answer 1

Hint: We start solving the problem by dividing the cubic equation \[a{{x}^{3}}+2b{{x}^{2}}+4cx+8d=0\] with 8 both sides. We then make necessary calculations and assume a variable for $\dfrac{x}{2}$ which makes the equation resembling \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\]. We then equate $\dfrac{x}{2}$ to all the roots 1, 5, 7 and make necessary calculations to get the required roots of \[a{{x}^{3}}+2b{{x}^{2}}+4cx+8d=0\].

Complete step-by-step answer:
According to the problem, we are given that 1, 5, 7 are the roots of \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\]. We need to find the roots of \[a{{x}^{3}}+2b{{x}^{2}}+4cx+8d=0\].
We have given \[a{{x}^{3}}+2b{{x}^{2}}+4cx+8d=0\]. Let us divide the whole equation with 8 on both sides.
So, we get \[\dfrac{a{{x}^{3}}+2b{{x}^{2}}+4cx+8d}{8}=\dfrac{0}{8}\].
$\Rightarrow \dfrac{a{{x}^{3}}}{8}+\dfrac{2b{{x}^{2}}}{8}+\dfrac{4cx}{8}+\dfrac{8d}{8}=0$.
$\Rightarrow \dfrac{a{{x}^{3}}}{8}+\dfrac{b{{x}^{2}}}{4}+\dfrac{cx}{2}+d=0$.
$\Rightarrow a{{\left( \dfrac{x}{2} \right)}^{3}}+b{{\left( \dfrac{x}{2} \right)}^{2}}+c\left( \dfrac{x}{2} \right)+d=0$.
Let us assume $\dfrac{x}{2}=y$. So, we get the cubic equation as $a{{y}^{3}}+b{{y}^{2}}+cy+d=0$. We can see that this cubic equation is similar to the cubic equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\].
So, the roots of the cubic equation $a{{y}^{3}}+b{{y}^{2}}+cy+d=0$ will be y = 1, 5, 7.
Now, we use $\dfrac{x}{2}=y$ to find the actual roots of \[a{{x}^{3}}+2b{{x}^{2}}+4cx+8d=0\].
So, we have $\dfrac{x}{2}=1\Leftrightarrow x=2$.
$\Rightarrow \dfrac{x}{2}=5\Leftrightarrow x=10$.
$\Rightarrow \dfrac{x}{2}=7\Leftrightarrow x=14$.
We have found the roots of the cubic equation \[a{{x}^{3}}+2b{{x}^{2}}+4cx+8d=0\] as 2, 10 and 14.

So, the correct answer is “Option (a)”.

Note: Whenever we get this type of problems, we try to convert the required equation into the equivalent form of the equation for which we already know the roots. We can also solve this problem as shown below.
We know that if $\alpha $, $\beta $ and $\gamma $ are the roots of the cubic equation $p{{x}^{3}}+q{{x}^{2}}+rx+s=0$, then
$\Rightarrow \alpha +\beta +\gamma =\dfrac{-q}{p}$.
$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{s}{p}$.
$\Rightarrow \alpha \beta \gamma =\dfrac{-t}{p}$.
Now, we have \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\Leftrightarrow {{x}^{3}}+\dfrac{b{{x}^{2}}}{a}+\dfrac{cx}{a}+\dfrac{d}{a}=0\].
So, we get $\dfrac{-b}{a}=1+5+7=13$, $\dfrac{c}{a}=\left( 1\times 5 \right)+\left( 5\times 7 \right)+\left( 7\times 1 \right)=5+35+7=47$ and $\dfrac{-d}{a}=\left( 1\times 5\times 7 \right)=35$.
So, the cubic equation will be ${{x}^{3}}-13{{x}^{2}}+47x-35=0$.
Now, we need to find the roots of \[a{{x}^{3}}+2b{{x}^{2}}+4cx+8d=0\Leftrightarrow {{x}^{3}}+\dfrac{2b{{x}^{2}}}{a}+\dfrac{4cx}{a}+\dfrac{8d}{a}=0\].
Let us substitute the values of $\dfrac{-b}{a}$, $\dfrac{c}{a}$ and $\dfrac{-d}{a}$. So, we get the cubic equation as \[{{x}^{3}}-26{{x}^{2}}+188x-280=0\].
$\Rightarrow {{x}^{3}}-24{{x}^{2}}-2{{x}^{2}}+140x+48x-280=0$.
$\Rightarrow {{x}^{3}}-24{{x}^{2}}+140x-2{{x}^{2}}+48x-280=0$.
$\Rightarrow x\left( {{x}^{2}}-24x+140 \right)-2\left( {{x}^{2}}-24x+140 \right)=0$.
$\Rightarrow \left( x-2 \right)\left( {{x}^{2}}-24x+140 \right)=0$.
$\Rightarrow \left( x-2 \right)\left( {{x}^{2}}-14x-10x+140 \right)=0$.
$\Rightarrow \left( x-2 \right)\left( x\left( x-14 \right)-10\left( x-14 \right) \right)=0$.
$\Rightarrow \left( x-2 \right)\left( x-10 \right)\left( x-14 \right)=0$.
$\Rightarrow x-2=0$, $x-10=0$ and $x-14=0$.
$\Rightarrow x=2$, $x=10$, $x=14$.
∴ The roots of \[a{{x}^{3}}+2b{{x}^{2}}+4cx+8d=0\] are 2, 10, 14.