Quadratic Polynomial Roots: Determining the Other Root

This solution explains how to find the unknown root of a quadratic polynomial, \(f(x) = ax^2 + bx + c\), given specific conditions related to its values and roots.

Understanding the Given Information

We are provided with a quadratic polynomial defined as \(f(x) = ax^2 + bx + c\). The following conditions are given:

• The value of the function at \(x=1\) is 2: \(f(1) = 2\)
• The value of the function at \(x=4\) is 2: \(f(4) = 2\)
• One root of the equation \(f(x) = 0\) is \(x = 2\).

Our objective is to determine the value of the other root of the equation \(f(x) = 0\).

Step-by-Step Solution Derivations

Step 1: Formulating Equations from Function Values

Using the given conditions \(f(1) = 2\) and \(f(4) = 2\), we can write two equations based on the polynomial definition \(f(x) = ax^2 + bx + c\):

1. \(f(1) = a(1)^2 + b(1) + c = a + b + c\). Since \(f(1) = 2\), we have: \(a + b + c = 2 \quad (1)\)
2. \(f(4) = a(4)^2 + b(4) + c = 16a + 4b + c\). Since \(f(4) = 2\), we have: \(16a + 4b + c = 2 \quad (2)\)

To simplify, let's subtract equation (1) from equation (2):

\((16a + 4b + c) - (a + b + c) = 2 - 2\)

\(15a + 3b = 0\)

We can rearrange this to express \(b\) in terms of \(a\):

\(3b = -15a\)

\(b = -5a \quad (3)\)

Step 2: Expressing 'c' in Terms of 'a'

Now, substitute the expression for \(b\) from equation (3) back into equation (1) (\(a + b + c = 2\)):

\(a + (-5a) + c = 2\)

\(-4a + c = 2\)

Solving for \(c\), we get:

\(c = 4a + 2 \quad (4)\)

Step 3: Utilizing the Known Root Information

We are given that \(x = 2\) is a root of the equation \(f(x) = 0\). This means that when we substitute \(x = 2\) into the polynomial, the result should be zero:

\(f(2) = a(2)^2 + b(2) + c = 4a + 2b + c\)

Therefore, we have the equation:

\(4a + 2b + c = 0 \quad (5)\)

Step 4: Solving for the Coefficient 'a'

Substitute the expressions for \(b\) (from equation 3) and \(c\) (from equation 4) into equation (5):

\(4a + 2(-5a) + (4a + 2) = 0\)

Simplify the equation:

\(4a - 10a + 4a + 2 = 0\)

\((4a - 10a + 4a) + 2 = 0\)

\(-2a + 2 = 0\)

Now, solve for \(a\):

\(2a = 2\)

\(a = 1\)

Step 5: Calculating the Coefficients 'b' and 'c'

With the value \(a = 1\), we can now find the values of \(b\) and \(c\) using equations (3) and (4):

• Using equation (3): \(b = -5a = -5(1) = -5\)
• Using equation (4): \(c = 4a + 2 = 4(1) + 2 = 4 + 2 = 6\)

Step 6: Constructing the Specific Quadratic Polynomial

Now we have all the coefficients: \(a = 1\), \(b = -5\), and \(c = 6\). The quadratic polynomial is:

\(f(x) = 1x^2 - 5x + 6\)

Which simplifies to:

\(f(x) = x^2 - 5x + 6\)

Step 7: Finding the Roots of the Polynomial Equation

To find the roots of \(f(x) = 0\), we need to solve the quadratic equation:

\(x^2 - 5x + 6 = 0\)

This equation can be solved by factoring:

We look for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3.

So, the factored form is:

\((x - 2)(x - 3) = 0\)

Setting each factor to zero gives the roots:

• \(x - 2 = 0\) => \(x = 2\)
• \(x - 3 = 0\) => \(x = 3\)

Final Conclusion on the Other Root

The roots of the quadratic equation \(f(x) = 0\) are \(x = 2\) and \(x = 3\). Since the question states that \(x = 2\) is one of the roots, the other root must be \(x = 3\).