Question

Any aryabatta here please help me only this question​

Answer 1

Answer:

Option C.) -6√3 < b < 6√3

Step-by-step explanation:

Given,

P(x) = 12x² - 2bx + 9

Here, a = 12 , b = -2 , c = 9

We know that, for any Quadratic equation,

Virtual roots(no real roots) = D < 0

Equal roots = D = 0

Unique and Distinct Roots = D > 0

Here, since, there is no real root, then,

D < 0

We know that,

D = b^2 - 4ac

So, by applying the value of D, we get,

b^2 - 4ac > 0

Let us now apply the values of each of these.

=> (-2)^2 - 4 × 12 × 9 < 0

=> 4 - 234 < 0

=> -230 < 0

Hence, the case satisfies, that P(x) has no real roots.

Now let us try trial and error method with the options,

• Option A,

b < -6√3

we know that, b = 2 , which is a positive number. And -6√3 is a negative number. Hence, 2 > -6√3 . So this option is wrong.

• Option B,

b > 6√3

We know that, 6√3 = 6 × 1.732 = 10.392

6√3 > b (b = 2)

So this option is also wrong.

• Option C,

This option is correct as -6√3 < b and b < 6√3. So, -6√3 < b < 6√3

(Proved above)

Hence, this is the right option.

• Option D,

This option is also wrong using the above results.

So, we get, the correct option as C.) -6√3 < b < 6√3

Answer 2

Answer:

here is your answer Hope it helps you