Question

Q.1
Write the sets in roster form:
{x:x^4-5x^2+6=0;x€R}​

Answer 1

Answer:

$\{ x | x^4 - 5x^2 + 6=0 , x \in R \} = \phi$

Step-by-step explanation:

The set in set-builder form is,

$A = \{ x | x^4 - 5x^2 + 6=0 , x \in R \}$

We can simplify the expression as,

$x^4 - 5x^2 + 6 = 0\\x^2( x^2 + 5 ) = 0\\x^2 + 5 = 0$

According to this expression,

$x^2 + 5 = 0\\x^2 = -5$

No real number satisfies above the equation. As we are given x belongs to Real Numbers.

Hence, this is a null set. It could be written as,

$\{ x | x^4 - 5x^2 + 6=0 , x \in R \} = \phi$