$${\text{Using }}2\left( { - \frac{1}{2}{x^2}} \right) = x\left( {{x^2} + 1} \right) + 6,\,{\text{we get }}{x^3} + {x^2} + x + 6 = 0.\,{\text{By trial }}x = - 2\,\,{\text{satisfies it}}{\text{.}}$$
$$\therefore \,\,\left( {x + 2} \right)\left( {{x^2} - x + 3} \right) = 0$$
$$ \Rightarrow \,\,x = - 2\,\,{\text{because }}{x^2} - x + 3 = 0\,\,{\text{has }}D < 0.$$
∴ the given terms are $$- 10, - 2, 6,$$   where common difference = 8.

$$\eqalign{ & {\text{If }}x = n \in Z,{n^2} + {\left( {n + 1} \right)^2} = 25\,\,\,{\text{or, }}2{n^2} + 2n - 24 = 0 \cr & {\text{or, }}{n^2} + n - 12 = 0 \cr & \therefore \,\,n = 3, - 4 \cr & \therefore \,\,x = 3, - 4 \cr & {\text{If }}x = n + k,n \in Z,0 < k < 1\,\,{\text{then }}{\left( {n + 1} \right)^2} + {\left( {n + 2} \right)^2} = 25 \cr & {\text{or, }}2{n^2} + 6n - 20 = 0\,\,{\text{or, }}{n^2} + 3n - 10 = 0 \cr & \therefore \,\,n = 2, - 5 \cr & \therefore \,\,x = 2 + k, - 5 + k,\,\,{\text{where }}0 < k < 1. \cr} $$

$$\eqalign{ & \alpha - 1 < 0,\beta - 1 > 0\,\,{\text{and }}D > 0 \cr & \Rightarrow \,\,\left( {\alpha - 1} \right)\left( {\beta - 1} \right) < 0\,\,{\text{and }}D > 0 \cr & \Rightarrow \,\,\alpha \beta - \left( {\alpha + \beta } \right) + 1 < 0\,\,{\text{and }}D > 0. \cr} $$

Let $$f\left( x \right) = {x^4} + a{x^3} + 3x - b.\,{\text{Then }}f\left( 1 \right) = 0,f\left( { - 1} \right) = 0.$$

$${x^2} - 4 \ne 0$$
$$ \Rightarrow \,\,x \ne 2.$$   But the given equation implies that if $$x \ne 2$$  then $$x = 2.$$  This is a contradiction, so $$x$$ has no value

$$\eqalign{ & {3^{x + 1}} = {2^{{{\log }_2}3}} \times {3^{{{\log }_2}3}} = 3 \times {3^{{{\log }_2}3}} \cr & \therefore \,\,{3^x} = {3^{{{\log }_2}3}}. \cr} $$

$${\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}} = 1$$
Case I
$${x^2} - 5x + 5 = 1\,\,{\text{and }}{x^2} + 4x - 60$$       can be any real number
$$ \Rightarrow \,\,x = 1,4$$
Case II
$${x^2} - 5x + 5 = - 1\,\,{\text{and }}{x^2} + 4x - 60$$       has to be an even number
$$ \Rightarrow \,\,x = 2,3$$
where 3 is rejected because for $$x = 3,{x^2} + 4x - 60$$    is odd.
Case III
$${x^2} - 5x + 5$$   can be any real number and $${x^2} + 4x - 60 = 0$$
$$ \Rightarrow \,\,x = - 10,6$$
⇒ Sum of all values of $$x = - 10 + 6 + 2 + 1 + 4 = 3$$

Let the second root be $$\alpha $$.
$$\eqalign{ & {\text{Then}}\,\,\alpha + \left( {1 - p} \right) = - p \cr & \alpha = - 1 \cr & {\text{Also}}\,\,\,\alpha .\left( {1 - p} \right) = 1 - p \cr & \Rightarrow \,\,\left( {\alpha - 1} \right)\left( {1 - p} \right) = 0 \cr & \Rightarrow \,\,p = 1\left[ {\because \,\alpha = - 1} \right] \cr & \therefore \,\,{\text{Roots are }}\alpha = - 1\,\,{\text{and}}\,\,p - 1 = 0 \cr} $$

Product of real root $$ = \frac{9}{{{t^2}}} > 0,\,\forall \,t \in R$$
∴ Product of real roots is always positive.

$$\eqalign{ & a{\left| x \right|^2} + 2b\left| x \right| - c = 0 \cr & \therefore \,\,\left| x \right| = \frac{{ - 2b \pm \sqrt {4{b^2} + 4ac} }}{2} = - b \pm \sqrt {{b^2} + ac} . \cr} $$
$$a, b, c$$  are positive. So, $$\left| x \right| = - b + \sqrt {{b^2} + ac} .$$
∴ $$x$$ has two real values.