Let α, β be roots of x²-5x+6=0. Find α³+β³-3αβ.
17
The quadratic equation is \(x^2 - 5x + 6 = 0\), with roots \(\alpha\) and \(\beta\).
By comparing coefficients, \(\alpha + \beta = 5\) and \(\alpha\beta = 6\).
Using the identity \(\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)\):
\(\alpha^3 + \beta^3 = 5^3 - 3(6)(5) = 125 - 90 = 35\).
So, \(\alpha^3 + \beta^3 - 3\alpha\beta = 35 - 3(6) = 35 - 18 = 17\).
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 + \(\frac{1}{x^3}\) .
