To find the quadratic equation given its roots, we use the relationship between the roots and the coefficients of the equation. If the roots of a quadratic equation are α and β, the equation can be written as: $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$ or $x^2 - (\alpha + \beta)x + \alpha\beta = 0$
The given roots are:
Calculate the sum and product using the given roots:
Substitute the calculated sum and product into the standard equation format:
$x^2 - (\alpha + \beta)x + \alpha\beta = 0$
$x^2 - (1)x + (-6) = 0$
$x^2 - x - 6 = 0$
This equation matches option 2.
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 ______.
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 + \(\frac{1}{x^3}\) .