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

The given equation can be written as
\({ 2 }^{ z }\left( 1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } \right) ={ 7 }^{ x }\left( 1+\frac { 1 }{ 7 } +\frac { 1 }{ 49 } \right) \)
\(\Rightarrow { 2 }^{ x }\left( \frac { 8+4+2 }{ 8 } \right) ={ 7 }^{ x }\left( \frac { 49+7+1 }{ 49 } \right) \)
\(\Rightarrow { 2 }^{ x }\left( \frac { 7 }{ 4 } \right) ={ 7 }^{ x }\left( \frac { 57 }{ 49 } \right) \Rightarrow \frac { 7 }{ 4 } \times \frac { 49 }{ 57 } =\frac { { 7 }^{ x } }{ { 2 }^{ x } } \)
\(\Rightarrow \frac { 7 }{ 4 } \times \frac { 49 }{ 57 } ={ \left( \frac { 7 }{ 4 } \right) }^{ x }\Rightarrow \frac { { 7 }^{ 3 } }{ 4\times 57 } ={ \left( \frac { 7 }{ 4 } \right) }^{ x }\)
\(\Rightarrow xlog\left( \frac { 7 }{ 4 } \right) =3log\ 7-log4-log57\)
\(\Rightarrow x=\frac { 3log7-log4-log57 }{ log\left( \frac { 7 }{ 2 } \right) } \)

Given sin (e^x) -5^x + 5^-x
We have \({ 5 }^{ x }+{ 5 }^{ -x }={ ({ 5 }^{ \frac { x }{ 2 } }-{ 5 }^{ \frac { -x }{ 2 } }) }^{ 2 }+2\ge 2\)
If sin (e^x) -5^x + 5^-x has a solution
We get sin (e^x) ≥ 2 which i not possible for any
real x as |sin 0|≤ 1 for all θ∈ R.
∴ Sin (e^x) = 5^X + 5^-x has no solution.

Given \((\sqrt { 5 } -\sqrt { 3 } )\) is a root
Another root \(\Rightarrow \sqrt { 5 } +\sqrt { 3 } \) 
∴ Sum of the roots \(=\sqrt { 5 } -\sqrt { 3 } +\sqrt { 5 } +\sqrt { 3 } =2\sqrt { 5 } \)
Product of the roots 
\(=(\sqrt { 5 } -\sqrt { 3 } )(\sqrt { 5 } +\sqrt { 3 } )\)
\(=(\sqrt { 5 } )^{ 2 }-{ (\sqrt { 3 } ) }^{ 2 }=5-3=2\)
∴ One of the factor is x² -x (sum of the roots) + product of the roots 
\(\Rightarrow { x }^{ 2 }-2x\sqrt { 5 } +2\)
The other factor also will be \({ x }^{ 2 }-2x\sqrt { 5 } +2\)
\(({ x }^{ 2 }-2x\sqrt { 5 } +2)({ x }^{ 2 }+2x\sqrt { 5 } +2)=0\)
\(\Rightarrow ({ x }^{ 2 }+2-2\sqrt { 5 } x)({ x }^{ 2 }+2+2\sqrt { 5 } x)=0\)
\(\Rightarrow \left( { x }^{ 2 }+2 \right) ^{ 2 }-{ (2\sqrt { 5 } x) }^{ 2 }=0\)
\([\because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 }]\)
\(\Rightarrow { x }^{ 4 }+{ 4x }^{ 2 }+4{ (5)x }^{ 2 }=0\)
\(\Rightarrow { x }^{ 4 }+{ 4x }^{ 2 }+4-{ 20x }^{ 2 }=0\)
\(\Rightarrow { x }^{ 4 }-{ 16x }^{ 2 }+4=0\) is a rational co-efficient polynomial equation.

Since \(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) is a root, x-\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) is a factor. To remove the outermost square root, we take x +\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) as another factor and find their product.
\(\left( x+\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \right) \left( x-\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \right) ={ x }^{ 2 }-\frac { \sqrt { 2 } }{ \sqrt { 3 } } \)
Still we didn’t achieve our goal. So we include another factor x²+\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) and get the product.
\(\left( { x }^{ 2 }-\frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) \left( { x }^{ 2 }+\frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) ={ x }^{ 4 }-\frac { 2 }{ 3 } \)
So, 3x⁴- 2 = 0 is a required polynomial equation with the integer coefficients.
Now we identify the nature of roots of the given equation without solving the equation. The idea comes from the negativity, equality to 0, positivity of Δ = b²- 4ac.

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

x⁸- 3x + 1 = 0
Here a_n = 1, a_o = 1
If \(\frac{p}{q}\) is a root of the polynomial, then as 
(p, q) = 1p is a factor of a_o = 1 and q is a factor of a_n = 1
Since 1 has no factors, the given equation has no rational roots.

Let  α, β and γ  be the roots of x³+ ax²+ bx + c = 0
Then, we get
Σ₁ = α + β + γ = -a ....(1)
Σ₂ = αβ + βγ + γα = b ...(2)
Σ₃ = αβγ = -c ...(3)
We have to form the equation whose roots are  α², β² and γ².
Using (1), (2) and (3), we find the following
Σ₁ = α² + β² + γ² = (α + β + γ )² - 2( αβ + βγ + γα) = (-a)² -2(b) = a²-2b,
Σ₂ = α²β² +  β²γ² + γ²α² = (αβ + βγ + γα)² - 2((αβ)( βγ)(γα) + (γα)(αβ))
= (αβ + βγ + γα)² - 2αβγ (β + γ + α) = (b)² - 2(-c)(-a) = b²-2ca
Σ₃ = α²β²γ² = (αβγ)² = (-c)² = c²
Hence, the required equation is
x³-(α² + β² + γ²)x² + (α²β² + β²γ² + γ²α²)x -  α²β²γ² = 0
That is, x³-(a²-2b)x² + (b²-2ca)x-c² = 0

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

Here Δ = b²−4ac = 0 for equal roots. This implies 4(k + 2)² = 4(9)k. This implies k = 4 or 1.

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