Question

Please help me on this problem

Answer 1

The pairs of integer having two real solution for
`ax^(2) -6x+c = 0` are

1. 
   `a = -4, c = 5`
2. 
   `a = 1, c = 6`
3. 
   `a = 2, c = 3`
4. 
   `a = 3, c = 3`

Explanation:

Given

`ax^(2) -6x+c = 0`

Now we will solve the equation by putting all the 6 pairs so we get the following

`-3x^(2) -6x-5 = 0` for
`a = -3 , c=-5`

`-4x^(2) -6x+5 = 0` for
`a = -4 , c=5`

`1x^(2) -6x+6 = 0` for
`a = 1 , c=6`

`2x^(2) -6x+3 = 0` for
`a = 2 , c=3`

`3x^(2) -6x+3 = 0` for
`a = 3 , c=3`

`5x^(2) -6x+4 = 0` for
`a = 5 , c=4`

The above all are Quadratic equations inn general form
`ax^(2) +bx+c=0`

where we have a,b and c constant values

So for a real Solution we must have

`Disciminant , b^(2) -4*a*c \geq 0`

for
`a = -3 , c=-5` we have

`Discriminant =-24` which is less than 0 ∴ not a real solution.

for
`a = -4 , c=5` we have

`Discriminant = 116` which is greater than 0 ∴ a real solution.

for
`a = 1 , c=6` we have

`Discriminant =12` which is greater than 0 ∴ a real solution.

for
`a = 2 , c=3` we have

`Discriminant =12` which is greater than 0 ∴ a real solution.

for
`a = 3 , c=3` we have

`Discriminant =0` which is equal to 0 ∴ a real solution.

for
`a = 5 , c=4` we have

`Discriminant =-44` which is less than 0 ∴ not a real solution.