Question

If x³+ax+1=0and x⁴+ax²+1=0 have common root then the exhaustive set of value of a is

Answer 1

For the given equations to share a common root, the exhaustive set for a is
`\( a \in (-\infty, -2] \cup [2, +\infty) \)`, where
`\( |a| \geq 2 \).`

Let's work through this:

Given equations:

1.
`\( x^3 + ax + 1 = 0 \)`

2.
`\( x^4 + ax^2 + 1 = 0 \)`

If these equations have a common root, let's say x = r, then substituting x = r into both equations should satisfy both.

1. For
`\( x^3 + ax + 1 = 0 \)`:

`\[ r^3 + ar + 1 = 0 \]`

2. For
`\( x^4 + ax^2 + 1 = 0 \)`:

`\[ r^4 + ar^2 + 1 = 0 \]`

Now, let's consider the discriminants of the above quadratic equations.

1. Discriminant of
`\( x^3 + ax + 1 = 0 \)` is
`\( a^2 - 4 \)`

2. Discriminant of
`\( x^4 + ax^2 + 1 = 0 \)` is
`\( a^2 - 4 \)`

For real roots, the discriminant must be greater than or equal to zero. Therefore:

`\[ a^2 - 4 \geq 0 \]`

Solving for a :

`\[ a^2 \geq 4 \\\\\[ |a| \geq 2 \]`

So, any real value of a such that |a|
`\geq` 2 would satisfy the condition that the two equations have a common root. The exhaustive set of values for a is
`\( (-\infty, -2] \cup [2, +\infty) \)`.