Assertion (A ) : the roots ` x^4 -5x^2 +6=0` are `+- sqrt(2) ,+-sqrt(3)`
Reason (R ) : the equation having the roots `alpha_1 , alpha_2 , ….., alpha_n` is ` (x-alpha_1) ( x- alpha_2 )…. (x-alpha_n)=0`

Options

A. both A and R are true R is the correct explanation of A
B. both A and R are true and R is not correct explanation of A
C. A is true and R is false
D. A is false and R is true

Answer

To solve the problem, we need to verify the assertion and the reason provided in the question.

### Step 1: Solve the equation \( x^4 - 5x^2 + 6 = 0 \)

We start by substituting \( x^2 = t \). This gives us:

\[ t^2 - 5t + 6 = 0 \]

### Step 2: Factor the quadratic equation

Next, we can factor the quadratic equation:

\[ (t - 3)(t - 2) = 0 \]

### Step 3: Find the values of \( t \)

Setting each factor to zero gives us:

\[ t - 3 = 0 \quad \Rightarrow \quad t = 3 \]
\[ t - 2 = 0 \quad \Rightarrow \quad t = 2 \]

### Step 4: Substitute back to find \( x \)

Since \( t = x^2 \), we have:

1. \( x^2 = 3 \) \( \Rightarrow x = \pm \sqrt{3} \)
2. \( x^2 = 2 \) \( \Rightarrow x = \pm \sqrt{2} \)

### Step 5: List the roots

Thus, the roots of the equation \( x^4 - 5x^2 + 6 = 0 \) are:

\[ x = \pm \sqrt{3}, \quad x = \pm \sqrt{2} \]

### Step 6: Verify the assertion

The assertion states that the roots are \( \pm \sqrt{2} \) and \( \pm \sqrt{3} \). Since we have found these roots, the assertion is true.

### Step 7: Verify the reason

The reason states that the equation having roots \( \alpha_1, \alpha_2, \ldots, \alpha_n \) can be expressed as:

\[ (x - \alpha_1)(x - \alpha_2) \ldots (x - \alpha_n) = 0 \]

This is indeed true for any polynomial equation. Therefore, the reason is also true.

### Conclusion

Both the assertion and the reason are true, and the reason correctly explains the assertion.

### Final Answer

Both assertion (A) and reason (R) are true, and R is the correct explanation for A.

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