The roots of the given equation $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$ are

$A$ prime number $p$ is called special if there exist primes $p_1, p_2, p_3, p_4$ such that $p = p_1 + p_2 = p_3 - p_4$. The number of special primes is

If $p$ and $q$ are prime numbers satisfying the condition $p^{2}-2q^{2}=1$,then the value of $p^{2}+2q^{2}$ is

If each root of the equation $2x^3 + ax^2 - 8x + b = 0$ is reduced by $1$,then in the transformed equation thus formed,the term containing $x^2$ and the constant term vanish. The roots of the original equation are

Let $a, b \in \mathbb{R}, a \neq 0$ be such that the equation $a x^{2}-2 b x+5=0$ has a repeated root $\alpha,$ which is also a root of the equation $x^{2}-2 b x-10=0$. If $\beta$ is the other root of this equation,then $\alpha^{2}+\beta^{2}$ is equal to:

If $a, b$ and $c$ are in arithmetic progression,then the roots of the equation $a x^{2}-2 b x+c=0$ are