Quadratic Equations

If the roots of $x^3+a x^2+b x+c=0$ are in arithmetic progression with common difference 1 , then

If the quadratic equations $3 x^2-7 x+2=0$ and $k x^2+7 x-3=0$ have a common root then the positive value of $k$ is

$p$ and $q$ are two roots of the equation $x^2+7 x+3=0$. If $\frac{3 p}{1-2 p}, \frac{3 q}{1-2 q}$ are the roots of $l x^2+m x+n=0$ and the greatest common divisor of $l, m, n$ is 1 , then $l-m+n=$

The maximum value of $\left\{x \in \mathbf{R} / \sqrt{x+2}>\sqrt{8-x^2}\right\}=$

If $\alpha, \beta$ are the roots of $a x^2+b x+c=0$ then $\left(\frac{\alpha}{a \beta+b}\right)^3-\left(\frac{\beta}{a \alpha+b}\right)^3=$

For $n>2$ and $n \in \mathbf{N}$, the product of the roots of $(x-n)\left(\left(x^2-2 n x\right)^2+\left(2 n^2-5\right)\left(x^2-2 n x\right)\right. \left.+\left(n^4-5 n^2+4\right)\right)=0$ is divisible by

If $\alpha_1, \beta_1, \gamma_1, \delta_1$ are the roots of the equation $a x^4+b x^3+c x^2+d x+e=0$ and $\alpha_2, \beta_2, \gamma_2, \delta_2$ are the roots of the equation $e x^4+d x^3+c x^2+b x+a=0$ such that $0<\alpha_1<\beta_1<\gamma_...

For the equation $x^4+x^3-4 x^2+x-1=0$ the ratio of the sum of the squares of all the roots to the product of the distinct roots is

$\alpha$ is the maximum value of $1-2 x-5 x^2$ and $\beta$ is the minimum value of $x^2-2 x+r$. If $5 \alpha x^2+\beta x+6>0$ for all real values $x$, then the interval in which $r$ lies is

Let $S$ be the set of all possible integral values of $\lambda$ in the interval $(-3,7)$ for which the roots of the quadratic equation $\lambda x^2+13 x+7=0$ are all rational numbers. Then the sum of the elements in $S$ is

$p$ is non-zero real number. If the equation whose roots are the squares of the roots of the equation $x^3-p x^2+p x-1=0$ is identical with the given equation, then $p=$

The minimum value of $\frac{9 \cdot 3^{2 x}+6 \cdot 3^x+4}{9 \cdot 3^{2 x}-6 \cdot 3^x+4}$ is

If $\alpha$ and $\beta$ are the real roots of the equation $\sqrt{\frac{5 x}{x-2}}+\sqrt{\frac{x-2}{5 x}}=\frac{29}{10}$ and $\alpha>\beta$, then $\sqrt{\alpha^2-11^4 \beta^2}=$