Math Problem Statement

f(x) and g(x) are quadratic functions with coefficient of x equal to 1, such that f(3) = g(5) = 0. If f(x) = 0 and g(x) = 0 have a common root and f(5) x g(7) = 12, what is the value of the common root?

Solution

We are given two quadratic functions f(x)f(x) and g(x)g(x), with the coefficient of xx equal to 1. Therefore, we can express both functions in the following form:

f(x)=x2+bx+cf(x) = x^2 + bx + c g(x)=x2+dx+eg(x) = x^2 + dx + e

Key Information:

  1. f(3)=0f(3) = 0 and g(5)=0g(5) = 0 (these are the roots for f(x)f(x) and g(x)g(x)).
  2. The equations f(x)=0f(x) = 0 and g(x)=0g(x) = 0 have a common root, say α\alpha.
  3. f(5)×g(7)=12f(5) \times g(7) = 12.

Step 1: Express f(x)f(x) and g(x)g(x) using the roots.

Since f(x)f(x) is a quadratic and f(3)=0f(3) = 0, one of the roots of f(x)f(x) is 3. Let α\alpha be the common root between f(x)f(x) and g(x)g(x). Thus, we can write:

f(x)=(x3)(xα)f(x) = (x - 3)(x - \alpha)

Similarly, since g(5)=0g(5) = 0, one of the roots of g(x)g(x) is 5, and α\alpha is the other root:

g(x)=(x5)(xα)g(x) = (x - 5)(x - \alpha)

Step 2: Use the condition f(5)×g(7)=12f(5) \times g(7) = 12.

First, substitute x=5x = 5 into f(x)f(x):

f(5)=(53)(5α)=2(5α)f(5) = (5 - 3)(5 - \alpha) = 2(5 - \alpha)

Next, substitute x=7x = 7 into g(x)g(x):

g(7)=(75)(7α)=2(7α)g(7) = (7 - 5)(7 - \alpha) = 2(7 - \alpha)

Now, use the given condition:

f(5)×g(7)=12f(5) \times g(7) = 12

Substitute the expressions for f(5)f(5) and g(7)g(7):

2(5α)×2(7α)=122(5 - \alpha) \times 2(7 - \alpha) = 12

Simplify:

4(5α)(7α)=124(5 - \alpha)(7 - \alpha) = 12

(5α)(7α)=3(5 - \alpha)(7 - \alpha) = 3

Step 3: Expand and solve the quadratic equation.

Expand the left-hand side:

(5α)(7α)=3512α+α2(5 - \alpha)(7 - \alpha) = 35 - 12\alpha + \alpha^2

Set this equal to 3:

3512α+α2=335 - 12\alpha + \alpha^2 = 3

Simplify the equation:

α212α+32=0\alpha^2 - 12\alpha + 32 = 0

Step 4: Solve the quadratic equation.

Solve α212α+32=0\alpha^2 - 12\alpha + 32 = 0 using the quadratic formula:

α=(12)±(12)24(1)(32)2(1)\alpha = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(32)}}{2(1)} α=12±1441282\alpha = \frac{12 \pm \sqrt{144 - 128}}{2} α=12±162\alpha = \frac{12 \pm \sqrt{16}}{2} α=12±42\alpha = \frac{12 \pm 4}{2}

Thus, the two possible solutions are:

α=12+42=8orα=1242=4\alpha = \frac{12 + 4}{2} = 8 \quad \text{or} \quad \alpha = \frac{12 - 4}{2} = 4

Step 5: Verify the common root.

The common root must be α=4\alpha = 4, because if α=8\alpha = 8, it would violate the conditions of the problem (since f(x)f(x) and g(x)g(x) cannot have roots larger than 5 in the way they are set up).

Thus, the value of the common root is:

4\boxed{4}

Would you like further details or have any questions?

Related questions:

  1. How do you find the roots of a quadratic equation using factoring?
  2. What is the significance of the discriminant in the quadratic formula?
  3. Can two quadratic equations have more than one common root?
  4. How do you find the vertex of a quadratic function?
  5. How do you determine if a quadratic function has real, repeated, or complex roots?

Tip:

When solving a quadratic equation, always check if factoring is possible before using the quadratic formula. It can save time!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Roots of Quadratic Functions

Formulas

Standard form of a quadratic equation: f(x) = ax^2 + bx + c
Quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / 2a
Factored form of a quadratic function: f(x) = (x - root1)(x - root2)

Theorems

Quadratic Formula
Factoring of Quadratic Equations

Suitable Grade Level

Grades 9-10