Question

The number of equations of the form

ax2

+ bx + 2 = 0 that can be formed if the equation

have real roots (a > 2, b < 6 and a and b are

positive integers) is/are

Answer 1

The discriminant of a quadratic equation D, is given by
D = b2 – 4ac 


It is determined from the coefficients of the equation ax2 + bx + c = 0. 

The value of D reveals what type of roots the equation has.


If D is greater than 0, we get real roots


For ax2 + bx + 2 = 0 


The discriminant = b2 – 4(a)(2)
                               = b2 – 8a


Equating this to 0
b2 – 8a = 0
8a = b2
a = b2/8


a and b are positive integers and b is less than 6

the values that b can take include: 1, 2, 3, 4, 5

when
b =1, a = 1/8


b =2, a = 4/8 = ½


b = 3, a = 9/8


b =4, a = 16/8 = 2


b = 5, a = 25/8

The only value of b that gives an integer value of a is 4


Meaning there is only one equation that can be formed that satisfy this condition

Answer 2

Answer:

6

Step-by-step explanation:

given equation

$ax^{2} + bx +2 = 0$

given

$a \geq 2$ and $b\leq 6$

finding discriminant of the equation

$b^{2} - 4ac\\b^{2} - 4(a)(2)\\b^{2} - 8a$

it is given that only for real roots of equation so

$b^{2} - 8a\geq 0\\b^{2}\geq 8a$

by input the least value of 'a' we can check the first value of 'b'

since least value of 'a' = 2

$b^{2}\geq 8(2)\\b^{2} \geq 16\\b\geq 4$

so possible values of b becomes 4,5,6

now when b=4, a= 2

$4^{2} \geq 8a\\16\geq 8a\\2\geq a\\a\leq 2\\a=2$(least value of a=2)

when b=5 , a= 2 , 3

$b^{2} \geq 8a\\5^{2} \geq 8a\\3\geq a\\ a\leq 3\\a=2,3$ (least value of a =2)

when b=6 , a=2,3,4

$b^{2} \geq 8a\\6^{2} \geq 8a\\36\geq 8a\\4\geq a\\a\leq 4\\a= 2,3,4$ (least value of a =2)

so total possible values of a and b becomes 6.

since for each different combination of a and b different equations will be formed.