What is the value of α (α ≠ 0) for which x 2– 5x + α and x 2– 7x + 2α have a common factor?
6
Finding Alpha for Common Factor in Quadratic Equations
The problem asks for the value of \(\alpha\) (\(\alpha \ne 0\)) such that two quadratic expressions, \(x^2 - 5x + \alpha\) and \(x^2 - 7x + 2\alpha\), have a common factor. This means that the two quadratic equations formed by setting these expressions to zero, namely \(x^2 - 5x + \alpha = 0\) and \(x^2 - 7x + 2\alpha = 0\), share a common root.
Let the common root be \(r\). If \(r\) is a root of both equations, it must satisfy both equations:
- \(r^2 - 5r + \alpha = 0\)
- \(r^2 - 7r + 2\alpha = 0\)
We have a system of two linear equations in terms of \(r^2\), \(r\), and \(\alpha\). We can eliminate \(r^2\) by subtracting the first equation from the second equation.
Subtracting equation (1) from equation (2):
\((r^2 - 7r + 2\alpha) - (r^2 - 5r + \alpha) = 0\)
\(r^2 - 7r + 2\alpha - r^2 + 5r - \alpha = 0\)
Combining like terms:
\((r^2 - r^2) + (-7r + 5r) + (2\alpha - \alpha) = 0\)
\(0 - 2r + \alpha = 0\)
This gives us a relationship between the common root \(r\) and \(\alpha\):
\(\alpha = 2r\)
Now substitute this expression for \(\alpha\) into the first equation (\(r^2 - 5r + \alpha = 0\)):
\(r^2 - 5r + (2r) = 0\)
\(r^2 - 3r = 0\)
We can factor this quadratic equation in \(r\):
\(r(r - 3) = 0\)
This equation yields two possible values for the common root \(r\):
- \(r = 0\)
- \(r - 3 = 0 \implies r = 3\)
Now we find the corresponding values of \(\alpha\) using the relationship \(\alpha = 2r\) for each possible value of \(r\).
Case 1: \(r = 0\)
If the common root is \(r=0\), then \(\alpha = 2(0) = 0\). However, the problem statement explicitly says that \(\alpha \ne 0\). Therefore, \(r=0\) is not the common root we are looking for under the given condition.
Case 2: \(r = 3\)
If the common root is \(r=3\), then \(\alpha = 2(3) = 6\). This value of \(\alpha\) is non-zero, which satisfies the condition \(\alpha \ne 0\).
Let's verify if \(\alpha=6\) results in the two quadratic expressions having a common factor. The equations become:
- Equation 1: \(x^2 - 5x + 6 = 0\)
- Equation 2: \(x^2 - 7x + 2(6) = x^2 - 7x + 12 = 0\)
Factoring the first equation:
\(x^2 - 5x + 6 = 0\)
\((x - 2)(x - 3) = 0\)
The roots are \(x=2\) and \(x=3\).
Factoring the second equation:
\(x^2 - 7x + 12 = 0\)
\((x - 3)(x - 4) = 0\)
The roots are \(x=3\) and \(x=4\).
Indeed, when \(\alpha=6\), both quadratic equations have a common root, which is \(x=3\). Therefore, the expressions \(x^2 - 5x + 6\) and \(x^2 - 7x + 12\) have a common factor, \((x-3)\).
Thus, the value of \(\alpha\) for which the two quadratic expressions have a common factor is 6.
Summary of Steps to Find Alpha
To find the value of \(\alpha\) for which the two quadratic polynomials share a common factor, we followed these steps:
- Assumed a common root, \(r\).
- Substituted the common root into both polynomial equations.
- Solved the resulting system of equations for \(r\) and \(\alpha\).
- Considered the condition \(\alpha \ne 0\).
- Identified the correct value of \(\alpha\).
| Step | Description |
|---|---|
| 1 | Assume common root \(r\): \(r^2 - 5r + \alpha = 0\), \(r^2 - 7r + 2\alpha = 0\) |
| 2 | Subtract equations: \((r^2 - 7r + 2\alpha) - (r^2 - 5r + \alpha) = 0 \implies -2r + \alpha = 0\) |
| 3 | Find \(\alpha\) in terms of \(r\): \(\alpha = 2r\) |
| 4 | Substitute \(\alpha\) back: \(r^2 - 5r + 2r = 0 \implies r^2 - 3r = 0\) |
| 5 | Solve for \(r\): \(r(r - 3) = 0 \implies r = 0\) or \(r = 3\) |
| 6 | Apply \(\alpha \ne 0\): \(r=0 \implies \alpha=0\) (rejected), \(r=3 \implies \alpha=6\) (accepted) |
| 7 | Final Value: \(\alpha = 6\) |
Revision Table: Key Concepts
| Concept | Explanation |
|---|---|
| Common Factor | If two polynomials have a common factor \((x-r)\), then \(x=r\) is a common root of the equations formed by setting the polynomials to zero. |
| Roots of a Quadratic | The roots of \(ax^2 + bx + c = 0\) are the values of \(x\) that satisfy the equation. These correspond to the factors of the quadratic expression. |
| System of Equations | Two or more equations involving the same variables. Solving the system finds the values of the variables that satisfy all equations simultaneously. |
Additional Information: Common Roots
When two polynomial equations share a common root, that root satisfies both equations. For quadratic equations \(a_1x^2 + b_1x + c_1 = 0\) and \(a_2x^2 + b_2x + c_2 = 0\), if they have a common root \(r\), then:
\(a_1r^2 + b_1r + c_1 = 0\)
\(a_2r^2 + b_2r + c_2 = 0\)
This creates a system of linear equations in terms of \(r^2\) and \(r\) (treating them as variables) and the coefficients \(a_1, b_1, c_1, a_2, b_2, c_2\). We can solve this system using methods like elimination or substitution to find the common root or a condition on the coefficients for a common root to exist.
In this specific problem, the coefficients were simple (\(a_1=a_2=1\)), making subtraction an easy way to eliminate the \(r^2\) term and find a direct relationship between \(r\) and \(\alpha\). This relationship was then used to find the possible values of \(r\), which in turn gave the possible values of \(\alpha\). The constraint \(\alpha \ne 0\) was crucial in selecting the correct value.
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