Question

Find the number of distinct real solutions of the system of equations $x+y+z=4;x^2+y^2+z^2=14;x^3+y^3+z^3=34$;

if $x^3-4x^2+x+6=0$ has three distinct roots.

• A. Greater than $3$
• B. Greater than $6$
• C. Less than $6$
• D. Equal to $6$.

Answer 1

Answer: (A), (D)

The correct options are
A Greater than $3$
D Equal to $6$.

Let $f\left(u\right)=a{u}^3+b{u}^3+cu+d$ be a $fnc$ of ${}^{\prime }{u}^{\prime }$.
If $x,y,z$ are roots of $f\left(u\right)=0$, Sum of Roots $=x+y+z=4$
$\Rightarrow$ But sum of Roots $\equiv -b/a$ ....(of $a{u}^3+b{u}^2+cu+d=0)$
So $-b/a=4\Rightarrow \frac{b}{a}=-4$
Now, $x^2+y^2+z^2=(x+y+z{)}^2-2(xy+yz+zx)$
$14=16-2$ (sum of Products of Roots taken two at a time)
$14=16-2\left(\frac{c}{a}\right)$
$\Rightarrow \frac{c}{a}=1$

Now, $x,y,z$ are roots of $f\left(u\right)=0$
$\Rightarrow f\left(x\right)=0;f\left(y\right)=0;f\left(z\right)=0$
ie $a{x}^3+b{x}^2+cx+d=0;a{y}^3+b{y}^2+cy+d=0;a{z}^3+b{z}^2+cz+d=0$
ie $x^3+\left(\frac{b}{a}\right)x^2+\left(\frac{c}{a}\right)x+\frac{d}{a}=0;y^3+\left(\frac{b}{a}\right)y^2+\left(\frac{c}{a}\right)y+\frac{d}{a}=0;z^3\left(\frac{b}{a}\right)z^2+\left(\frac{c}{a}\right)z+\frac{d}{a}=0$

$\Rightarrow$ Adding these three equations, gives$(x^3+y^3+z^3)+\left(\frac{b}{a}\right)(x^2+y^2+z^2)+\left(\frac{c}{a}\right)(x+y+z)+\frac{3d}{a}=0+0+0=0$
$\Rightarrow 34+(-4)\left(14\right)+\left(2\right)\left(4\right)+\frac{3d}{a}=0$
$\Rightarrow \frac{3d}{a}=18\Rightarrow \frac{d}{a}=6$
So $f\left(u\right)=a{u}^3+b{u}^2+cu+d=0$
$\Rightarrow \frac{1}{a}f\left(u\right)=u^3+\left(\frac{b}{a}\right)u^2+\left(\frac{c}{a}\right)u+\frac{d}{a}=0$
$\Rightarrow u^3-4u^2+u+6=0$

Now, $u^3-4u^2+u+6=0$ has $3$ distinct roots (as is given)
Let they be $=\alpha ,\beta ,\gamma$

So, $x=\alpha ,y=\beta ,z=\gamma$ is a solution.
But, by symmetry of the variables $x,y,z$; we can write five more permutations for

$x$ ,	$y$ ,	$z$.
$\alpha ,$ $\alpha ,$ $\beta ,$ $\beta ,$ $\gamma ,$ $\gamma ,$	$\beta ,$ $\gamma ,$ $\alpha ,$ $\gamma ,$ $\alpha ,$ $\beta ,$	$\gamma \to solnNo.1$ $\beta ,\to solnNo.2$ $\gamma \to solnNo.3$ $\alpha \to solnNo.4$ $\beta \to solnNo.5$ $\alpha \to solnNo.6$

So, $6$ distinct real solutions possible.