Find an equation...?

Question

find an equation that has the following solutions : 1/10 and 3/2

Answer 1

You could find it for a quadratic equation:
ax^2 + bx + c where x = (-b +/- (b^2 - 4ac))/2a

Change the values for x so they have the same denominator.
x = 2/20 and 30/20

Use the denominator of both sides of the equation to get:
2a = 20
a = 10

Back to our formula:
x = (-b +/- (b^2 - 4ac))/2a
2/20, 30/20 = (-b +/- (b^2 - 4ac))/2a
2, 30 = -b +/- (b^2 - 4(10)c)
2, 30 = -b +/- (b^2 - 40c)

rewrite 2,30 as 16 +/- 14
16 +/- 14 = -b +/- (b^2 - 40c)

Solve one side of the +/- to find b:
-b = 16
b = -16

Now find c from the other side:
14 = (b^2 - 40c)
14 = 256 - 40c
40c = 256 - 14 = 242
c = 6.05

Formula: ax^2 + bx + c
10x^2 - 16x + 6.05

Check:
-b +/- (b^2 - 4ac) / 2a
16 +/- (256 - 242) / 20
16/20 +/- 14/20
x = 2/20, 30/20 = 1/10, 3/2 (check!)