The quadratic equation that has the roots 5/3 and 3/5 is (x - 5/3)(x - 3/5) = 0

=> x^2 - 3x/5 - 5x/3 + 1 = 0

=> x^2 - 34x/15 + 1 = 0

=> 15x^2 - 34x + 15 = 0

**The required quadratic equation is 15x^2 - 34x + 15 = 0**

We'll use Viete's relations and we'll form the quadratic when we know the sum and the product of the roots:

x1 + x2 = 5/3 + 3/5

x1 + x2 = (25 + 9)/15

x1 + x2 = 34/15

S = x1 + x2 = 34/15

x1*x2 = (5/3)*(3/5)

x1*x2 = 1

**P = 1**

We'll form the quadratic, knowing the values of the sum and the product:

x^2 - Sx + P = 0

x^2 - 34x/15 + 1 = 0

We'll multiply by 15 and we'll get the final form of the quadratic whose sum is 34/15 and product is 1.

15x^2 - 34x + 15 = 0

We also could use the fact that a quadratic equation could be written as a product of linear factors, when the solutions are given.

ax^2 + bx + c = (x - 5/3)(x -3/5)

We'll remove the brackets;

ax^2 + bx + c = x^2 - 3x/5 - 5x/3 + 1

**We'll combine like terms and we'll get:**

**ax^2 + bx + c = x^2 - 34x/15 + 1**