The positive value of m for which the roots of the equation ${12}{x}^2 + mx + 6 = 0$ are in the ratio of 2 : 3 is ______.
Finding Positive m Value for Ratio Roots
We are given the quadratic equation ${12}{x}^2 + mx + 6 = 0$. The goal is to find the positive value of m when the equation's roots are in the ratio 2:3.
Representing Roots by Ratio
Let the roots of the equation be $2\alpha$ and $3\alpha$, satisfying the given $2:3$ ratio.
Applying Vieta's Formulas
For a quadratic equation $ax^2 + bx + c = 0$, Vieta's formulas state:
- Sum of roots = $-\frac{b}{a}$
- Product of roots = $\frac{c}{a}$
Applying these to ${12}{x}^2 + mx + 6 = 0$:
- Sum: $2\alpha + 3\alpha = 5\alpha = -\frac{m}{12}$
- Product: $(2\alpha)(3\alpha) = 6\alpha^2 = \frac{6}{12} = \frac{1}{2}$
Calculating Root Multiplier α
From the product of roots:
$6\alpha^2 = \frac{1}{2}$Solving for $\alpha^2$:
$\alpha^2 = \frac{1}{12}$Therefore, $\alpha$ can be:
$\alpha = \pm \sqrt{\frac{1}{12}} = \pm \frac{1}{2\sqrt{3}}$Determining the Value of m
From the sum of roots, $5\alpha = -\frac{m}{12}$. We can express $m$ in terms of $\alpha$:
$m = -60\alpha$Substitute the values of $\alpha$ to find $m$:
- Case 1: $\alpha = \frac{1}{2\sqrt{3}}$ $m = -60 \left( \frac{1}{2\sqrt{3}} \right) = -\frac{30}{\sqrt{3}} = -\frac{30\sqrt{3}}{3} = -10\sqrt{3}$
- Case 2: $\alpha = -\frac{1}{2\sqrt{3}}$ $m = -60 \left( -\frac{1}{2\sqrt{3}} \right) = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3}$
Since the question asks for the positive value of m, the answer is $10\sqrt{3}$.
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Important Questions from Quadratic Equation
If 2x 2+ 5x + 1 = 0, then one of the values of \(x - \frac{1}{{2x}}\) is:
If \(a-\frac{12}{a}=1\) , where a > 0, then the value of \(a^2+\frac{16}{a^2}\) is:
If x 2 – 3x + 1 = 0, then the value of \(\frac{(x^4+\frac{1}{x^2})}{(x^2+5x+1)}\) is:
If \(\sqrt{x}{}-{1\over\sqrt{x}}=\sqrt5\) , \(x \ne 0\) , then what is the value of \((x^4+{1\over{x^2}})\over(x^2+1) \) ?
If x 2 + \(\frac{1}{x^2}\) = 18, x > 0, then find the value of x 3 + \(\frac{1}{x^3}\) .
