Nature of Roots

Let $a, b, c$ be the sides of a triangle. No two of them are equal and $\lambda \in R$. If the roots of the equation $x^2+2\left(a+b+c\right)x+3\lambda \left(ab+bc+ca\right)=0$ are real, then

The smallest value of $k$, for which both the roots of the equation, $x^2-8kx+16\left(k^2-k+1\right)=0$ are real, distinct and have value atleast $4$, is

Let $p\left(x\right)$ be a quadratic polynomial with real coefficients. If $p\left(x\right)=0$ has only purely imaginary roots, then the zeroes of the polynomial $p\left(p\left(x\right)\right)$ are

If the roots of the quadratic equation $x^2+8x+{\mu }^2+6\mu =0$ are real, then $\mu$ lies between

Define a function $f\left(x\right)=\frac{16x^2-96x+153}{x-3}$ for all real $x\ne 3$. The least positive value of $f\left(x\right)$ is

The number of integral values of $m$ for which the equation $\left(1+m^2\right)x^2-2\left(1+3m\right)x+\left(1+8m\right)=0$ has no real root, is

The sum of the abscissae of the points where the curves, $y=k{x}^2+(5k+3)x+6k+5, (k\in R)$ touch the $X$-axis, is equal to

If $2+i\sqrt{3}$ is a root of the equation $x^2+px+q=0$, then the value of $(p, q)$ is

If the equation $x^2- \left(2+m\right)x+\left(m^2-4m+4\right)=0$ has coincident roots, then-

If c + 1 < b, then the roots of the quadratic equation x² - bx + c = 0 is 

The number of real roots of the equation ${\left(\mathrm{x}\mathrm{ }-\mathrm{ }1\right)}^2+{\left(\mathrm{x}\mathrm{ }-\mathrm{ }2\right)}^2+{\left(\mathrm{x}\mathrm{ }-\mathrm{ }3\right)}^2=0$ is -

How many ordered pairs of integers $(x, y)$ satisfy the equation $x^2+y^2=2(x+y)+xy?$

For $x\in \mathrm{ℝ},$ the given expression $\frac{x^2+34x-71}{x^2+2x-7}$, cannot lie between $a$ and $b.$ Find $a+b$.

Prove that if the roots of the equation $\left(a^4+b^4\right)x^2+4abcdx+\left(c^4+d^4\right)=0$ be real, then they cannot be unequal.

If the roots of the equation $x^2-2\left(a+b\right)x+a\left(a+2b+c\right)=0$ be equal, prove that $a,b,c$ are in geometric progression.

If the roots of the equation $a\left(b-c\right)x^2+b\left(c-a\right)x+c\left(a-b\right)=0$ are equal, then prove that $\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$.