Quadratic EquationHard

Question

If both roots of the equation $x^{2} - 2(a - 1)x + 2a + 1 = 0$ are positive, then

Options

A.$a < 2$
B.$a \geq 4$
C.$1 \leq a \leq 4$
D.$1 < a < 2$

Solution

$D \geq 0 \Rightarrow a \in ( - \infty,0\rbrack \cup \lbrack 4,\infty)$

$$\begin{matrix} \frac{b}{2a} > 0 \Rightarrow a - 1 > 0 \Rightarrow a \in (1,\infty) \\ f(0) > 0 \Rightarrow 2a + 1 > 0 \Rightarrow a \in \left( - \frac{1}{2},\infty \right) \end{matrix}$$

Hence, $a \in \lbrack 4,\infty)$

Create a free account to view solution

View Solution Free
Topic: Quadratic Equation·Practice all Quadratic Equation questions

More Quadratic Equation Questions

The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has...If the difference between the roots of the equation x2 + ax + 1 = 0 is less than &#8730;5, then the set of possible valu...The number log2 7 is...Both the roots of the equation (x - b)(x - c) + (x - a)(x - c) + (x - a)(x - b) = 0 are always...Given that the solution set of the quadratic inequality $ax^{2} + bx + c &gt; 0$ is $(2,3)$. Then the solution set of th...