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
\(\\ \frac { -3+\sqrt { 7 } }{ 2 } ,\frac { -3-\sqrt { 7 } }{ 2 } ,\frac { 2+\sqrt { 2 } }{ 2 } ,\frac { 2-\sqrt { 2 } }{ 2 } \)

Let ∝, β be the roots x²+ bx + ca = 0
∝ +β = -b, ∝β = ca
and a, ૪ be the roots of x² + cx + ab = 0
∝+૪ = -c, ∝૪ = ab
Then, a² + b∝ + ca = 0 and
∝²+ c∝ + ab = 0
⇒ (b-c)∝+a(c-b) = 0 ⇒ ∝ = a
Also ∝β = ca and ∝૪ = ab
∴ β = c and ૪ = c
since ∝ = a is a root of x² + bx + ca = 0,
we get a² + ba + ca = 0 ⇒ a + b + c = 0
Thus, β+૪ = b + c = -a and β૪ = bc
Hence β, ૪ are the roots of the equation
x² + ax + bc = 0

Given \(\frac { p }{ 2x } =\frac { a }{ x+x } +\frac { b }{ x-c } \)
\(\frac { p }{ 2x } =\frac { (a+b)x+c(-a) }{ { x }^{ 2 }-{ c }^{ 2 } } \)
⇒ P (x² - c²) = 2 (a + b) x² - 2c(a-b)x
⇒ (2a + 2b - p)x2 - 2c (a - b)x +pc² = 0
This equation has equal roots
if b²-4ac = 0
⇒ c²(a-b)² - pc² (2a + 2b - b) = 0
⇒ (a - b)² - 2p (a + b) +p² = 0 [∵ c² ≠ 0]
⇒ [p - (a + b)]² = (a + b)² - (a - b)² = 4ab
⇒ p-(a+b) = 土2\(\sqrt{ab}\)
⇒ p-(a+b)土2\(\sqrt{ab}\) = (\(\sqrt{a}\) 土\(\sqrt{b}\))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

\(\alpha +\beta =\frac { 1 }{ { \alpha }^{ 2 } } +\frac { 1 }{ { \beta }^{ 2 } } =\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 } }{ { \alpha }^{ 2 }{ \beta }^{ 2 } } =\frac { { (\alpha +\beta ) }^{ 2 }-2\alpha \beta }{ { \alpha }^{ 2 }{ \beta }^{ 2 } } \)
\(\Rightarrow \frac { -b }{ a } =\frac { \frac { { b }^{ 2 } }{ { a }^{ 2 } } -2\frac { c }{ a } }{ \frac { { c }^{ 2 } }{ { a }^{ 2 } } } =\frac { { b }^{ 2 }-2ac }{ { c }^{ 2 } } \)
\(\Rightarrow \frac { 2a }{ c } =\frac { { b }^{ 2 } }{ { c }^{ 2 } } +\frac { b }{ a } \)
\(\frac { 2a }{ c } =\frac { { ab }^{ 2 }+{ bc }^{ 2 } }{ { ac }^{ 2 } } \)
\(2{ a }^{ 2 }c={ ab }^{ 2 }+{ b }^{ 2 }\)
\(\frac { 2a }{ b } =\frac { b }{ c } +\frac { c }{ a } \) [Dividing by abc]
\(\Rightarrow \frac { c }{ a } ,\frac { a }{ b } ,\frac { b }{ c } \) are in A.P \(\Rightarrow \frac { a }{ c } ,\frac { b }{ a } ,\frac { c }{ b } \) are in H.P