Quadratic equation at x=0

[mpx86]

Homework Statement

A functions is defined as f(x) = ax2 + bx + c, where a, b, c are real numbers. If f(3) = f(– 2) = 0, what is the value of f(0)?

Homework Equations

The Attempt at a Solution

As function is 0 at 3, -2, therefore
9a + 3b + c=0 also,
4a -2 b + c=0
c=-6a or c= 6b
f(0)=c= 6a or -6b
So the answer depends on the value of a or b (a= -b)

However, the book states the following solution

f(x) = ax2 + bx + c and f(3) = f(– 2) = 0 implies that 3 and (–2) are the roots of f(x).
So, the quadratic equation is
f(x) = (x – 3)(x + 2) = x2 – x – 6 = 0.
Thus, f(0) = –6.

However, if the roots are 3, -2 to a quadratic equ. , then required equation should be
f(x) = k(x – 3)(x + 2)
Am I right?

 

[pasmith]

Homework Statement

A functions is defined as f(x) = ax2 + bx + c, where a, b, c are real numbers. If f(3) = f(– 2) = 0, what is the value of f(0)?

Homework Equations

The Attempt at a Solution

As function is 0 at 3, -2, therefore
9a + 3b + c=0 also,
4a -2 b + c=0
c=-6a or c= 6b
f(0)=c= 6a or -6b
So the answer depends on the value of a or b (a= -b)

However, the book states the following solution

f(x) = ax2 + bx + c and f(3) = f(– 2) = 0 implies that 3 and (–2) are the roots of f(x).
So, the quadratic equation is
f(x) = (x – 3)(x + 2) = x2 – x – 6 = 0.
Thus, f(0) = –6.

However, if the roots are 3, -2 to a quadratic equ. , then required equation should be
f(x) = k(x – 3)(x + 2)
Am I right?

Yes. Unless the question tells you what $a$ is, there is no way to uniquely determine f(0) from the knowledge that f(3) = f(-2) = 0.

 

[Ray Vickson]

Homework Statement

A functions is defined as f(x) = ax2 + bx + c, where a, b, c are real numbers. If f(3) = f(– 2) = 0, what is the value of f(0)?

Homework Equations

The Attempt at a Solution

As function is 0 at 3, -2, therefore
9a + 3b + c=0 also,
4a -2 b + c=0
c=-6a or c= 6b
f(0)=c= 6a or -6b
So the answer depends on the value of a or b (a= -b)

However, the book states the following solution

f(x) = ax2 + bx + c and f(3) = f(– 2) = 0 implies that 3 and (–2) are the roots of f(x).
So, the quadratic equation is
f(x) = (x – 3)(x + 2) = x2 – x – 6 = 0.
Thus, f(0) = –6.

However, if the roots are 3, -2 to a quadratic equ. , then required equation should be
f(x) = k(x – 3)(x + 2)
Am I right?

You are right, but you should note that k = a; that is, f(x) = a(x-3)(x+2).