Put 5+2√6 = y; Then 5-2√6 = \(\frac{1}{y}\)
∴ The given equataion becomes, 
\({ y }^{ { x }^{ 2 }-3 }+{ \left( \frac { 1 }{ y } \right) }^{ { x }^{ 2 }-3 }=10\)
Put  \({ y }^{ { x }^{ 2 }-3 }=z ...(1)\)
\(\Rightarrow z+\frac { 1 }{ z } =10\Rightarrow { z }^{ 2 }-10z+1=0\)
\(\Rightarrow z=\frac { 10\pm \sqrt { 100-4 } }{ 2 } =5\pm 2\sqrt { 6 } \)
\(\therefore (5+2\sqrt { 6 } )^{ { x }^{ 2 }-3 }=5\pm 2\sqrt { 6 } ={ (5\pm 2\sqrt { 6 } ) }^{ \pm 1 } \)
⇒x²- 3 = 1 or x²- 3 = -1
⇒ x² = 4 or x² = 2
⇒ x = 土2 or x = 土 √2
∴ The roots are 2,  -2, √2, -√2

Since x²- y² is even, either both x and y are even or both x and y are odd.
In any case both (x + y) and (x - y) are even integers
∴ x² - y²= (x + y) (x +y) must be divisible by 4 But 4 does not divide 353702
∴ x² - y² = 353702 has no positive integral solutions.

Let p(-x) = x⁹- 5x⁵ + 4x⁴ + 2x² + 1 = 0
P(x) has only one sign change. It has atmost one positive roots.
Also p(-x) = (-x)⁹ - 5(-x)⁵ + 4(-x)4 +2(-x)² +1 = 0
p(-x) = -x⁹+ 5x⁵+4x⁴+ 2x²+1 = 0
It has only one sign change.
It has atmost one negative roots.
Clearly 0 is not a root.
So maximum number of real roots is 3 and hence there are atleast six imaginary solutions.

Since 2-\(\sqrt{3}\)i is a root and the coefficients are rational numbers, 2+\(\sqrt{3}\)i  is also a root. A required polynomial equation is given by
x² −(Sum of the roots) x + Product of the roots = 0
and hence
x²- 4x +1 = 0 is a required equation.

Let α, β, γ and δ be the roots of ax⁴+ bx³+ cx²+ dx + e = 0
Σ₁ = α + β + γ + δ = -\(\frac { b }{ a } \),
Σ₂ = αβ + αγ + αδ + βγ + βδ + γδ = \(\frac { c }{ a } \),
Σ₃ = αβγ + αβδ +  αγδ +  βγδ = -\(\frac { d }{ a } \)
Σ₄ = αβγδ =\(\frac { e }{ a } \)
We have to find α² + β² + γ² + δ² 
Applying the algebraic identity
(a+b+c+d)² ≡ a²+b²+c²+d²+2(ab+ac+ad+bc+bd+cd),
we get
α² + β² + γ² + δ² = (α + β + γ + δ)²-2(αβ + αγ + αδ + βγ + βδ + γδ)
= \(\left( \frac { b }{ a } \right) ^{ 2 }-2\left( \frac { c }{ a } \right) \)
= \(\frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \).

Given equation is
(2x² - 3x + 1) (2x² + 5x + 1) = 9x² ....(1)
Clearly x = 0 does not satisfy (1),
∴ (1) can be rewritten as  
\(\left( 2x+\frac { 1 }{ x } -3 \right) \left( 2x+\frac { 1 }{ x } +5 \right) =9...(2)\)
put  \(2x+\frac { 1 }{ x } =y\)
∴ (2)⇒ (y - 3) (y + 5) = 9
⇒ y²+ 2y-15 = 9
or y²+ 2y - 24 = 0
⇒ (y + 6) (y - 4) = 0
⇒  y = -6, -4
Case (i)
When \(y=-6,2x+\frac { 1 }{ x } =-6\)
2x²+6x+1 = 0
\(\Rightarrow =\frac { -6\pm \sqrt { 36-8 } }{ 4 } \)
\(x=-3\pm \frac { \sqrt { 7 } }{ 2 } \)
Case(ii)
When \(y=4,2x+\frac { 1 }{ x } =4\)
⇒ 2x²- 4x+1 = 0
\(x=\frac{4 \pm \sqrt{16-8}}{4}\)
\(x=\frac { 2\pm \sqrt { 2 } }{ 2 } \)
This, the roots are
\(\\ \frac { -3+\sqrt { 7 } }{ 2 } ,\frac { -3-\sqrt { 7 } }{ 2 } ,\frac { 2+\sqrt { 2 } }{ 2 } ,\frac { 2-\sqrt { 2 } }{ 2 } \)

Given equation is (a²+b²+c²) p²-2
(ab+bc+cd) p+(b²+c²+d²) ≤ 0...(1)
(1) can be rewritten as
(a²p² - 2abp + b²) + (b²p² - 2bcp + c²) + (c²p2 - 2cdp + d2) ≤ 0
Since a, b, c, d, p∈ R
(ap-b)² ≥ 0, (bp-c)² ≥ 0 and (cp-d)² ≥ 0
∴ (2) will be satisfied only if
ap - b = 0, bp - c = 0, cp - d = 0
\(\Rightarrow \frac { b }{ a } =\frac { c }{ b } =\frac { d }{ c } =p\)
⇒ a, b, c, d are in G.P and ad = bc

Here, a = 9, b = - 36, c = 44, d = -16
Since the roots form an arithmetic progression,
Let the roots be a - d, a and a + d
Sum of the roots \(=\frac { -b }{ a } \)
\(\Rightarrow (a-d)+(a)+(a+d)=\frac { -(-36) }{ 9 } =4\)
\(\Rightarrow 3a=4\Rightarrow a=\frac { 4 }{ 3 } \)
and product of the roots \(=\frac { -d }{ a } \)
\(=\frac { -(-16) }{ 9 } =\frac { 16 }{ 9 } \)
\(\Rightarrow (a-d)(a)(a+d)=\frac { 16 }{ 9 } \)
\(\left( { a }^{ 2 }-{ d }^{ 2 } \right) (a)=\frac { 16 }{ 9 } \)
\(\left( \frac { 16 }{ 9 } -{ d }^{ 2 } \right) \left( \frac { 4 }{ 3 } \right) =\frac { 16 }{ 9 } \ [\because a=\frac { 4 }{ 3 } ]\)
\(\frac { 16 }{ 9 } -{ d }^{ 2 }=\frac { 16 }{ 9 } \times \frac { 3 }{ 4 } =\frac { 4 }{ 3 } \)
\(\frac { 16 }{ 9 } -\frac { 4 }{ 3 } ={ d }^{ 2 }=\frac { 16 }{ 9 } \times \frac { 3 }{ 4 } =\frac { 4 }{ 3 } \)
\(\frac { 16 }{ 9 } -\frac { 4 }{ 3 } ={ d }^{ 2 }\)
\(\Rightarrow { d }^{ 2 }=\frac { 16-12 }{ 9 } =\frac { 4 }{ 9 } \)
\(\Rightarrow d=\pm \sqrt { \frac { 4 }{ 9 } } =\frac { 2 }{ 3 } \)
∴ The roots are a-d, a, a+d
\(\Rightarrow \frac { 4 }{ 3 } -\frac { 2 }{ 3 } ,\frac { 4 }{ 3 } ,\frac { 4 }{ 3 } +\frac { 2 }{ 3 } \Rightarrow \frac { 2 }{ 3 } ,\frac { 4 }{ 3 } ,2\)

The given equation is same as
(2x-3)(3x-2)(6x-1)(x-2)-5 = 0
After a computation, the above equation becomes
(6x²-13x+6)(6x²-13x+12)-5 = 0
By taking y = 6x²-13x, the above equation becomes
(y+6)(y+12)-5 = 0
which is same as
y²+18y+7 = 0
Solving this equation, we get y = −1 and y = −7.
Substituting the values of y in y = −6x²-13x, we get
6x²-13x+1 = 0
6x²-13x+7 = 0
Solving these two equations, we get
x = 1, x = \(\frac { 7 }{ 6 } \), x = \(\frac { 13 + \sqrt { 145 } }{ 12 } \) and x = \(\frac { 13-\sqrt { 145 } }{ 12 } \)
as the roots of the given equation.

We can solve this fourth degree equation by rewriting it suitably and adopting a technique of substitution. Rewriting the equation as
(x−2)(x−3)(x−7)(x+2)+19 = 0
the given equation becomes
(x²−5x+6)(x²−5x−14)+19 = 0 .
If we take x² − 5x as y, then the equation becomes (y+6)(y−14)+19 = 0;
that is, y²-8y-65 = 0
Solving this we get solutions y = 13 and y = −5. Substituting this we get two quadratic equations
x²-5x-13 = 0 and x²-5x+5 = 0
which can be solved by usual techniques. The solutions obtained for these two equations together give solutions as \(\frac { 5\pm \sqrt { 77 } }{ 2 } ,\frac { 5\pm \sqrt { 5 } }{ 2 } \).

Since two roots are in the ratio p : q : r, we can assume the roots as pλ, qλ and rλ .
Then, we get
Σ₁ = pλ + qλ + rλ = -a .....(1)
Σ₂ = (pλ)(qλ)+(qλ)(rλ)+(rλ)(pλ) ..........(2)
Σ₃ = (pλ)(qλ)(rλ) = -c .....(3)
Now, we get
(1) ⇒ λ = -\(\frac { a }{ p+q+r } \) ..........(4)
(3) ⇒ λ³ = \(\frac { c }{ pqr } \) ...........(5)
Substituting (4) in (5), we get
\(\left( \frac { a }{ p+q+r } \right) ^{ 3 }=-\frac { c }{ pqr } \) ⇒ pqra³ = c(p+q+ r)³.

We have \(x=\sqrt { 2+x } \)
\(\Rightarrow { x }^{ 2 }=2+x \Rightarrow { x }^{ 2 }-x-2=0\)
\(\Rightarrow x=\frac { 1\pm \sqrt { 1+8 } }{ 2 } \Rightarrow x=\frac { 1\pm 3 }{ 2 } \)
\(\Rightarrow x=\frac { 1+3 }{ 2 } ,\frac { 1-3 }{ 2 } \Rightarrow x=2,-1\)
Also x>0, we get x = 2

Let ∝, β are the roots of the equation
Sum of the roots \(\alpha +\beta =\frac { -b }{ a } \)
 \(=\frac { [-(a-2)] }{ 1 } =a-2\)
Product of the roots \(=\alpha \beta =\frac { c }{ a } \)
\(=\frac { -(a+1) }{ 1 } =-(a+1)\)
we have \({ \alpha }^{ 2 }{ \beta }^{ 2 }=({ \alpha +\beta ) }^{ 2 }-2\alpha \beta \)
\(={ (a-2) }^{ 2 }+2(a+1)\)
\(={ a }^{ 2 }-4a+4+2a+2\)
\(=(a-1{ ) }^{ 2 }+5\)
Thus \(\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }\) is least if a = 1

Δ = b²- 4ac = -20 < 0. The roots are imaginary numbers.