The roots of the quadratic equation are 5/3 and 3/5

For a quadratic equation with roots x1 and x2, the equation can be written as (x - x1)*(x - x2) = 0

Therefore we have (x - 5/3)(x - 3/5) = 0

Make the denominator in the brackets the same

=> [(3x - 5)/3][(5x - 3)/5] = 0

Eliminate the denominator

=> (3x - 5)(5x - 3) = 0

Open the brackets

=> 15x^2 - 25x - 9x + 15 =0

Simplify

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

**Therefore the quadratic equation is 15x^2 - 34x + 15 =0**

If x1 and x2 are the roots of a quadratic equation, then (x-x1)(x=x2) = 0 is the quadratic equation.

Or x^2-(x1+x2)x+x1x2 = 0 is the quadratic equation

Or a{x^2+-(x1+x2)x+x1x2} = 0 is also the quadratic equation with x1 and x2 as the roots , where a is any constan.t

So if 5/3 and 3/5 are the roots, then

(x-3/5)((x-3/5) = 0 is the quadratic with roots 5/3 and 3/5 .

x^2-(5/3+3/5)x +(5/3)(3/5) = 0

x^2- (25+9)x/15 +1 = 0

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

15x^2- 34 +15 = 0 is the equation with (5/3) and (3/5) as roots.

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**