Question

The number of all possible positive integral values of alpha for which the roots of the quadratic equation , 6x square - 11x+ alpha =0 are rational numbers is

Answer 1

Answer:

All possible positive value of α is 3 , 4 and 5.

Step-by-step explanation:

Given Quadratic Equation,

6x² - 11x + α = 0

To find: All possible value of α so that roots are rational number.

We know that nature of roots of the Quadratic equation depends on the value of discriminant that is D = b² - 4ac

If D < 0 then roots are imaginary.

if D ≥ 0 then roots are real.

Also if D is not a perfect square then roots are irrational numbers and if D is a perfect square number that is a number whose square root exit then roots are rational number.

Here, a = 6 , b = -11 and c = α

D = (-11)² - 4ac = 121 - 4 × 6 × α = 121 - 24α

Now, For D to be perfect square value of α should be 3 , 4 , 5

since,

α = 3 ,

D = 121 - 24×3 = 121 - 72 = 49

and root of 49 = ± 7

α = 4 ,

D = 121 - 24×4 = 121 - 96 = 25

and root of 25 = ± 5

α = 5 ,

D = 121 - 24×5 = 121 - 120 = 1

and root of 1 = ± 1

Therefore, All possible positive value of α is 3 , 4 and 5.

Answer 2

Step-by-step explanation:

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