If the equation $\frac{{{x^2} + 5}}{2} = x - 2\cos \left( {ax + b} \right)$ has atleast one solution, then $(b + a)$ can be equal to

If $x$ is a solution of the equation, $\sqrt {2x + 1}  - \sqrt {2x - 1}  = 1, \left( {x \ge \frac{1}{2}} \right)$ , then $\sqrt {4{x^2} - 1} $ is equal to 

One root of the following given equation $2{x^5} - 14{x^4} + 31{x^3} - 64{x^2} + 19x + 130 = 0$ is

Let $\alpha$ and $\beta$ be the roots of $x^2-6 x-2=0$, with $\alpha>\beta$. If $a_n=\alpha^n-\beta^n$ for $n \geq 1$, then the value of $\frac{a_{10}-2 a_8}{2 a_9}$ is

Let $a, b, c$ be non-zero real roots of the equation $x^3+a x^2+b x+c=0$. Then,

Let $\alpha, \beta$ be two roots of the equation $x^{2}+(20)^{\frac{1}{4}} x+(5)^{\frac{1}{2}}=0$. Then $\alpha^{8}+\beta^{8}$ is equal to: