If $\alpha , \beta$ and $\gamma$ are the roots of ${x^3} + 8 = 0$, then the equation whose roots are ${\alpha ^2},{\beta ^2}$ and  ${\gamma ^2}$ is

If $x$ is real and satisfies $x + 2 > \sqrt {x + 4} ,$ then

Let $f(x)=x^4+a x^3+b x^2+c$ be a polynomial with real coefficients such that $f(1)=-9$. Suppose that $i \sqrt{3}$ is a root of the equation $4 x^3+3 a x^2+2 b x=0$, where $i=\sqrt{-1}$. If $\alpha_1, \alpha_2, \alpha_3$, and $\alpha_4$ are all the roots of the equation $f(x)=0$, then $\left|\alpha_1\right|^2+\left|\alpha_2\right|^2+\left|\alpha_3\right|^2+\left|\alpha_4\right|^2$ is equal to. . . . . .

The solution of the equation $2{x^2} + 3x - 9 \le 0$ is given by

Let $a, b, c, d$ be real numbers between $-5$ and $5$ such that  $|a|=\sqrt{4-\sqrt{5-a}},|b|=\sqrt{4+\sqrt{5-b}},|c|=\sqrt{4-\sqrt{5+c}}$ $|d|=\sqrt{4+\sqrt{5+d}}$ Then, the product $a b c d$ is

If $x$ be real, then the maximum value of $5 + 4x - 4{x^2}$ will be equal to