Question

Let $P\left(x\right)=x^4-8x^2+21$ and $Q\left(y\right)=y^2-6y+10,x,y\in R$. If $P\left(x\right).Q\left(y\right)=5$, and $x,y\in I$, then possible value(s) of $x+y$ is/are

• A. $1$
• B. $3$
• C. $5$
• D. $7$

Answer 1

Answer: (A), (C)

$1$
$5$

Looking at the equations carefully, we can find a possibility, that $P\left(x\right)=5$ and $Q\left(y\right)=1$
Thus, $x^4-8x^2+16=0,\Rightarrow x=2,-2$
And, $y^2-6y+9=0,\Rightarrow y=3$
Hence, the possible value of $x+y$ are $1$ and $5$.