# The quadratic equation has roots x=2/3 and x=-4. Find one set of possible values for a,b,c?

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Let f(x) = ax^2 + bx + c be a quadratic function.

Given the roots of f(x) are:

x1= 2/3

x2= -4

Then, ( x- x1) and ( x- x2) are factors of f(x).

Then, f(x) = ( x- 2/3) ( x+4)

or we can write: f(x) = ( 3x-2) ( x+ 4) such that 2/3 and -4 are the roots for f(x).

We will expand the brackets:

==> f(x) = 3x^2 - 2x + 12x - 8

==> f(x) = 3x^2 +10x - 8

Now we will compare the fucntions:

3x^2 + 10x - 8 = ax^2 + bx + C

Then, we conclude that:

**a= 3 b= 10 c = -8**

We'll put the quadratic as ax^2 + bx + c = 0

Now, we'll write the quadratic as a product of linear factors:

ax^2 + bx + c = (x - 2/3)(x+4)

We'll remove the brackets from the right side:

ax^2 + bx + c = x^2 + 4x - 2x/3 - 8/3

We'll compare both sides and we'll get:

**a = 1**

b = 4 - 2/3

**b = 10/3**

**c = -8/3**

Wealso could use Viete's relations to determine the coefficients a,b,c:

x1 + x2 = -b/a

But, from enunciation, x1 = 2/3 and x2 = -4:

2/3 - 4 = -b/a

We'll multiply by 3:

-10/3 = -b/a

**b/a = 10/3**

10a = 3b

x1*x2 = c/a

**-8/3 = c/a**

-8a = 3c

**The quadratic equation is:**

**ax^2 + bx + c = x^2 + 10x/3 - 8/3**