What is the positive value of m for which the roots of the equation 12x 2+ mx + 5 = 0 are in the ratio 3 ∶ 2?
5 √10
Understanding the Problem: Finding 'm' for Quadratic Roots Ratio
The question asks us to find the positive value of 'm' for the quadratic equation \(12x^2 + mx + 5 = 0\), given that its roots are in the ratio 3:2.
A quadratic equation in the standard form is \(ax^2 + bx + c = 0\). In our case, we have:
- \(a = 12\)
- \(b = m\)
- \(c = 5\)
Let the two roots of the equation be \(\alpha\) and \(\beta\).
Relating Roots to Coefficients
For any quadratic equation \(ax^2 + bx + c = 0\), there are standard formulas relating the roots (\(\alpha\) and \(\beta\)) to the coefficients (a, b, and c):
- Sum of roots: \(\alpha + \beta = -\frac{b}{a}\)
- Product of roots: \(\alpha \cdot \beta = \frac{c}{a}\)
Using the coefficients from our equation \(12x^2 + mx + 5 = 0\):
- Sum of roots: \(\alpha + \beta = -\frac{m}{12}\)
- Product of roots: \(\alpha \cdot \beta = \frac{5}{12}\)
Using the Given Ratio of Roots
We are given that the roots are in the ratio 3:2. This means we can write the relationship between the roots as:
\(\frac{\alpha}{\beta} = \frac{3}{2}\)
From this ratio, we can express one root in terms of the other. Let's express \(\alpha\) in terms of \(\beta\):
\(\alpha = \frac{3}{2}\beta\)
Solving for the Roots using Product and Ratio
We can substitute the expression for \(\alpha\) into the product of roots equation:
\(\alpha \cdot \beta = \frac{5}{12}\)
\(\left(\frac{3}{2}\beta\right) \cdot \beta = \frac{5}{12}\)
\(\frac{3}{2}\beta^2 = \frac{5}{12}\)
Now, we can solve for \(\beta^2\):
\(\beta^2 = \frac{5}{12} \times \frac{2}{3}\)
\(\beta^2 = \frac{10}{36}\)
\(\beta^2 = \frac{5}{18}\)
Taking the square root of both sides gives us the possible values for \(\beta\):
\(\beta = \pm\sqrt{\frac{5}{18}}\)
\(\beta = \pm\frac{\sqrt{5}}{\sqrt{18}}\)
Simplify the denominator \(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\):
\(\beta = \pm\frac{\sqrt{5}}{3\sqrt{2}}\)
To rationalize the denominator, multiply the numerator and denominator by \(\sqrt{2}\):
\(\beta = \pm\frac{\sqrt{5} \times \sqrt{2}}{3\sqrt{2} \times \sqrt{2}} = \pm\frac{\sqrt{10}}{3 \times 2} = \pm\frac{\sqrt{10}}{6}\)
So, the possible values for \(\beta\) are \(\frac{\sqrt{10}}{6}\) and \(-\frac{\sqrt{10}}{6}\).
Solving for 'm' using Sum and Ratio
Now we can use the sum of roots equation: \(\alpha + \beta = -\frac{m}{12}\)
Substitute \(\alpha = \frac{3}{2}\beta\):
\(\frac{3}{2}\beta + \beta = -\frac{m}{12}\)
\(\left(\frac{3}{2} + 1\right)\beta = -\frac{m}{12}\)
\(\frac{5}{2}\beta = -\frac{m}{12}\)
Solve for 'm':
\(m = -\frac{5}{2}\beta \times 12\)
\(m = -30\beta\)
Finding the Positive Value of 'm'
We have two possible values for \(\beta\): \(\frac{\sqrt{10}}{6}\) and \(-\frac{\sqrt{10}}{6}\). Let's find the corresponding values of 'm' for each case:
- If \(\beta = \frac{\sqrt{10}}{6}\):
\(m = -30 \times \left(\frac{\sqrt{10}}{6}\right)\)
\(m = -5\sqrt{10}\)
- If \(\beta = -\frac{\sqrt{10}}{6}\):
\(m = -30 \times \left(-\frac{\sqrt{10}}{6}\right)\)
\(m = 5\sqrt{10}\)
The question asks for the positive value of m.
Comparing the two values we found, the positive value of m is \(5\sqrt{10}\).
Summary of Steps
Here's a quick look at the steps we followed to solve this problem:
| Step | Description | Formula/Equation Used |
|---|---|---|
| 1 | Identify coefficients a, b, c | \(ax^2 + bx + c = 0\) |
| 2 | Write sum and product of roots in terms of coefficients | \(\alpha + \beta = -b/a\), \(\alpha \cdot \beta = c/a\) |
| 3 | Express root ratio mathematically | \(\alpha / \beta = 3/2\) |
| 4 | Use product of roots and ratio to find \(\beta^2\) | \(\alpha \cdot \beta = c/a\) with \(\alpha = (3/2)\beta\) |
| 5 | Solve for possible values of \(\beta\) | \(\beta = \pm \sqrt{\beta^2}\) |
| 6 | Use sum of roots and ratio to find 'm' in terms of \(\beta\) | \(\alpha + \beta = -b/a\) with \(\alpha = (3/2)\beta\) |
| 7 | Substitute \(\beta\) values to find 'm' values | \(m = -30\beta\) |
| 8 | Select the positive value of 'm' | Final answer |
Revision Table: Key Quadratic Concepts
| Concept | Description | Formula |
|---|---|---|
| Standard Form of Quadratic Equation | An equation of degree 2 | \(ax^2 + bx + c = 0\) (where \(a \neq 0\)) |
| Roots of a Quadratic Equation | The values of x that satisfy the equation (also called solutions or zeros) | |
| Sum of Roots | The sum of the two roots \(\alpha\) and \(\beta\) | \(\alpha + \beta = -b/a\) |
| Product of Roots | The product of the two roots \(\alpha\) and \(\beta\) | \(\alpha \cdot \beta = c/a\) |
| Discriminant | Helps determine the nature of the roots | \(\Delta = b^2 - 4ac\) |
Additional Information: Handling Root Ratios
When the roots of a quadratic equation \(ax^2 + bx + c = 0\) are in the ratio p:q, you can denote the roots as \(p\lambda\) and \(q\lambda\) for some constant \(\lambda\). This can sometimes simplify calculations.
Using this approach for the given problem, the roots could be \(3k\) and \(2k\) for some \(k\).
- Sum of roots: \(3k + 2k = 5k = -m/12\)
- Product of roots: \((3k)(2k) = 6k^2 = 5/12\)
From the product equation:
\(6k^2 = 5/12\)
\(k^2 = \frac{5}{12 \times 6} = \frac{5}{72}\)
\(k = \pm\sqrt{\frac{5}{72}} = \pm\frac{\sqrt{5}}{\sqrt{72}} = \pm\frac{\sqrt{5}}{6\sqrt{2}} = \pm\frac{\sqrt{10}}{12}\)
Now, substitute the value of \(k\) into the sum equation:
\(5k = -m/12\)
\(m = -12 \times 5k = -60k\)
Using \(k = \frac{\sqrt{10}}{12}\):
\(m = -60 \times \frac{\sqrt{10}}{12} = -5\sqrt{10}\)
Using \(k = -\frac{\sqrt{10}}{12}\):
\(m = -60 \times \left(-\frac{\sqrt{10}}{12}\right) = 5\sqrt{10}\)
Again, the positive value of m is \(5\sqrt{10}\). This confirms the result obtained earlier and shows an alternative method for handling root ratios.
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