If (x + 2) is a common factor of x 2+ ax + b and x 2+ bx + a, then the ratio of a ∶ b is equal to
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Finding the Ratio of Coefficients Using the Factor Theorem
The problem states that the expression \((x + 2)\) is a common factor of two quadratic polynomials: \(x^2 + ax + b\) and \(x^2 + bx + a\). We are asked to find the ratio of the coefficients \(a\) and \(b\), represented as \(a \ratio b\).
According to the Factor Theorem, if \((x - c)\) is a factor of a polynomial \(P(x)\), then \(P(c) = 0\). In this case, the common factor is \((x + 2)\), which can be written as \((x - (-2))\). Therefore, if \((x + 2)\) is a factor of a polynomial, substituting \(x = -2\) into the polynomial must result in zero.
Applying the Factor Theorem to the First Polynomial
Let the first polynomial be \(P(x) = x^2 + ax + b\). Since \((x + 2)\) is a factor, we have:
\[P(-2) = (-2)^2 + a(-2) + b = 0\] \[4 - 2a + b = 0\]Rearranging this equation, we get our first relationship between \(a\) and \(b\):
\[2a - b = 4 \quad \cdots (1)\]Applying the Factor Theorem to the Second Polynomial
Let the second polynomial be \(Q(x) = x^2 + bx + a\). Since \((x + 2)\) is also a factor of this polynomial, we have:
\[Q(-2) = (-2)^2 + b(-2) + a = 0\] \[4 - 2b + a = 0\]Rearranging this equation, we get our second relationship between \(a\) and \(b\):
\[a - 2b = -4 \quad \cdots (2)\]Solving the System of Linear Equations
Now we have a system of two linear equations with two variables, \(a\) and \(b\):
- \(2a - b = 4\)
- \(a - 2b = -4\)
We can solve this system using methods like substitution or elimination. Let's use the elimination method. Multiply equation (2) by 2:
\[2(a - 2b) = 2(-4)\] \[2a - 4b = -8 \quad \cdots (3)\]Now, subtract equation (3) from equation (1):
\[(2a - b) - (2a - 4b) = 4 - (-8)\] \[2a - b - 2a + 4b = 4 + 8\] \[3b = 12\]Solving for \(b\):
\[b = \frac{12}{3}\] \[b = 4\]Now substitute the value of \(b\) back into either equation (1) or (2) to find \(a\). Using equation (1):
\[2a - (4) = 4\] \[2a - 4 = 4\] \[2a = 4 + 4\] \[2a = 8\]Solving for \(a\):
\[a = \frac{8}{2}\] \[a = 4\]Determining the Ratio a : b
We found that \(a = 4\) and \(b = 4\). The ratio \(a \ratio b\) is therefore:
\[a \ratio b = 4 \ratio 4\]This ratio can be simplified by dividing both sides by the greatest common divisor, which is 4:
\[a \ratio b = \frac{4}{4} \ratio \frac{4}{4}\] \[a \ratio b = 1 \ratio 1\]Conclusion
If \((x + 2)\) is a common factor of \(x^2 + ax + b\) and \(x^2 + bx + a\), then the ratio of \(a\) to \(b\) is \(1 \ratio 1\). This means \(a\) is equal to \(b\).
| Polynomial | Condition from Factor Theorem | Equation Derived |
|---|---|---|
| \(x^2 + ax + b\) | Substitute \(x=-2\) | \(4 - 2a + b = 0 \implies 2a - b = 4\) |
| \(x^2 + bx + a\) | Substitute \(x=-2\) | \(4 - 2b + a = 0 \implies a - 2b = -4\) |
Revision Table: Key Concepts Reviewed
| Concept | Description | Application in Problem |
|---|---|---|
| Factor Theorem | If \((x - c)\) is a factor of \(P(x)\), then \(P(c) = 0\). | Used to set \(x = -2\) in both polynomials. |
| Root of a Polynomial | A value \(c\) such that \(P(c) = 0\). If \((x-c)\) is a factor, then \(c\) is a root. | \(x = -2\) is a common root for both polynomials. |
| Simultaneous Equations | A set of equations with multiple variables, solved together to find values that satisfy all equations. | The two equations derived from the Factor Theorem were solved to find \(a\) and \(b\). |
| Ratio | A comparison of two quantities. Represented as \(a \ratio b\) or \(\frac{a}{b}\). | Calculated the ratio of the determined values of \(a\) and \(b\). |
Additional Information: Understanding Polynomial Factors
A factor of a polynomial is an expression that divides the polynomial exactly, leaving no remainder. The Factor Theorem provides a direct link between the roots of a polynomial and its linear factors. If you know a root \(c\), you know \((x-c)\) is a factor, and vice versa.
In this problem, since \((x+2)\) is a common factor, it means \(x=-2\) is a common root of both quadratic equations \(x^2 + ax + b = 0\) and \(x^2 + bx + a = 0\). This property is what allows us to substitute \(x=-2\) and form equations to solve for the unknown coefficients \(a\) and \(b\). Common factors play an important role in simplifying polynomial expressions and finding their roots.
Understanding the relationship between factors, roots, and coefficients is fundamental in algebra. Problems like this reinforce the application of key theorems such as the Factor Theorem and techniques for solving systems of equations.
Important Questions from Compound Ratios
The train fare, bus fare and air fare between 2 places are in the ratio 5 : 8 : 12, the number of passenger travelled by them is in the ratio 3 : 4 : 5 and the total fare collected on a particular day for these modes of transportation for a single trip is Rs. 1,07,000. Find the fare collected from the air passengers.
A person carries Rs. 165/ - in the form of currency notes of denominations Rs. 5, Rs. 10 & Rs. 20 in the ratio of 3 : 2 : 1. What is the value of currency notes of Rs. 20 denomination?
If a: b = 5: 3, then (a³-b³): (a³+b³) = ?
What is the compound ratio of 2 : 3, 4 : 7 and 5 : 6?
If $X : Y = 1/2 : 1/5$ and $Y : Z = 1/3 : 1/4$, then find the ratio of $1/X : 1/Y : 1/Z$
