If one root of (a 2– 5a + 3)x 2+ (3a – 1)x + 2 = 0 is twice the other, then what is the value of ‘a’?
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Understanding the Quadratic Equation Problem
The question asks us to find the value of the variable 'a' in a given quadratic equation. The equation is \((a^2 - 5a + 3)x^2 + (3a - 1)x + 2 = 0\). We are also given a condition about the roots of this equation: one root is exactly twice the other root. To solve this, we will use the properties of the roots of a quadratic equation, specifically the sum and product of roots.
Identifying Coefficients and Root Relationship
A standard quadratic equation is in the form \(Ax^2 + Bx + C = 0\), where A, B, and C are coefficients. In our equation, \((a^2 - 5a + 3)x^2 + (3a - 1)x + 2 = 0\), we can identify the coefficients:
- Coefficient of \(x^2\), \(A = a^2 - 5a + 3\)
- Coefficient of \(x\), \(B = 3a - 1\)
- Constant term, \(C = 2\)
Note that for this to be a quadratic equation, the coefficient of \(x^2\) must not be zero, i.e., \(A \neq 0\).
Let the roots of the equation be \(\alpha_1\) and \(\alpha_2\). The problem states that one root is twice the other. Let's assume: \(\alpha_1 = \alpha\) \(\alpha_2 = 2\alpha\)
Using Properties of Roots: Sum and Product
For a quadratic equation \(Ax^2 + Bx + C = 0\), the sum of the roots is given by \(-\frac{B}{A}\), and the product of the roots is given by \(\frac{C}{A}\). We will apply these properties to our equation with roots \(\alpha\) and \(2\alpha\).
Sum of roots: \(\alpha + 2\alpha = 3\alpha = -\frac{B}{A}\) Substituting the values of A and B: \[3\alpha = -\frac{3a - 1}{a^2 - 5a + 3} \quad \text{(Equation 1)}\]
Product of roots: \(\alpha \times 2\alpha = 2\alpha^2 = \frac{C}{A}\) Substituting the values of A and C: \[2\alpha^2 = \frac{2}{a^2 - 5a + 3} \quad \text{(Equation 2)}\]
Solving for 'a' using the Root Equations
We now have two equations involving \(\alpha\) and 'a'. We can use these to find the value of 'a'. From Equation 2, we can simplify and express \(\alpha^2\) in terms of 'a': \(2\alpha^2 = \frac{2}{a^2 - 5a + 3}\) Dividing by 2: \[\alpha^2 = \frac{1}{a^2 - 5a + 3} \quad \text{(Equation 3)}\]
Now, let's square Equation 1 to get an expression involving \(\alpha^2\): \((3\alpha)^2 = \left(-\frac{3a - 1}{a^2 - 5a + 3}\right)^2\) \(9\alpha^2 = \frac{(3a - 1)^2}{(a^2 - 5a + 3)^2} \quad \text{(Equation 4)}\)
Now, substitute the expression for \(\alpha^2\) from Equation 3 into Equation 4: \(9 \left(\frac{1}{a^2 - 5a + 3}\right) = \frac{(3a - 1)^2}{(a^2 - 5a + 3)^2}\)
Assuming \(a^2 - 5a + 3 \neq 0\), we can multiply both sides by \((a^2 - 5a + 3)^2\): \(9 (a^2 - 5a + 3) = (3a - 1)^2\)
Now, expand and simplify the equation to solve for 'a': \(9a^2 - 45a + 27 = (3a)^2 - 2(3a)(1) + 1^2\) \(9a^2 - 45a + 27 = 9a^2 - 6a + 1\)
Subtract \(9a^2\) from both sides: \(-45a + 27 = -6a + 1\)
Gather the 'a' terms on one side and the constant terms on the other: \(27 - 1 = -6a + 45a\) \(26 = 39a\)
Solve for 'a': \(a = \frac{26}{39}\)
Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 13: \(a = \frac{26 \div 13}{39 \div 13}\) \(a = \frac{2}{3}\)
Verification
We should verify that for \(a = \frac{2}{3}\), the coefficient \(A = a^2 - 5a + 3\) is not zero, otherwise the original equation is not quadratic. \(A = \left(\frac{2}{3}\right)^2 - 5\left(\frac{2}{3}\right) + 3\) \(A = \frac{4}{9} - \frac{10}{3} + 3\) To combine these terms, find a common denominator, which is 9: \(A = \frac{4}{9} - \frac{10 \times 3}{3 \times 3} + \frac{3 \times 9}{1 \times 9}\) \(A = \frac{4}{9} - \frac{30}{9} + \frac{27}{9}\) \(A = \frac{4 - 30 + 27}{9}\) \(A = \frac{1}{9}\) Since \(A = \frac{1}{9} \neq 0\), the equation is indeed quadratic for \(a = \frac{2}{3}\), and our solution is valid.
Thus, the value of 'a' is \(\frac{2}{3}\).
Revision Table: Key Concepts
| Concept | Description | Formula for \(Ax^2 + Bx + C = 0\) |
|---|---|---|
| Quadratic Equation | An equation of the form \(Ax^2 + Bx + C = 0\), where \(A \neq 0\). | \(Ax^2 + Bx + C = 0\) |
| Roots of a Quadratic Equation | The values of the variable (usually \(x\)) that satisfy the equation. | Let roots be \(\alpha_1, \alpha_2\). |
| Sum of Roots | The sum of the roots is related to the coefficients B and A. | \(\alpha_1 + \alpha_2 = -\frac{B}{A}\) |
| Product of Roots | The product of the roots is related to the coefficients C and A. | \(\alpha_1 \times \alpha_2 = \frac{C}{A}\) |
Additional Information: Solving Quadratic Equations
Quadratic equations are fundamental in algebra. They can be solved using various methods:
- Factoring: If the quadratic expression can be factored into two linear expressions, setting each factor to zero gives the roots.
- Completing the Square: This method involves manipulating the equation to form a perfect square trinomial on one side, making it easier to isolate the variable.
- Quadratic Formula: The roots of \(Ax^2 + Bx + C = 0\) can always be found using the formula: \[x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\] The term \(B^2 - 4AC\) is called the discriminant, which tells us about the nature of the roots (real, complex, equal).
Understanding the relationship between the roots and the coefficients (Vieta's formulas) is crucial for problems like the one discussed, where specific conditions about the roots are given.
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