\(\alpha +\beta =\frac { 1 }{ { \alpha }^{ 2 } } +\frac { 1 }{ { \beta }^{ 2 } } =\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 } }{ { \alpha }^{ 2 }{ \beta }^{ 2 } } =\frac { { (\alpha +\beta ) }^{ 2 }-2\alpha \beta }{ { \alpha }^{ 2 }{ \beta }^{ 2 } } \)
\(\Rightarrow \frac { -b }{ a } =\frac { \frac { { b }^{ 2 } }{ { a }^{ 2 } } -2\frac { c }{ a } }{ \frac { { c }^{ 2 } }{ { a }^{ 2 } } } =\frac { { b }^{ 2 }-2ac }{ { c }^{ 2 } } \)
\(\Rightarrow \frac { 2a }{ c } =\frac { { b }^{ 2 } }{ { c }^{ 2 } } +\frac { b }{ a } \)
\(\frac { 2a }{ c } =\frac { { ab }^{ 2 }+{ bc }^{ 2 } }{ { ac }^{ 2 } } \)
\(2{ a }^{ 2 }c={ ab }^{ 2 }+{ b }^{ 2 }\)
\(\frac { 2a }{ b } =\frac { b }{ c } +\frac { c }{ a } \) [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

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 the coefficients of the equation are all rational numbers, 2+i and 3-\(\sqrt{2}\) are roots, we get 2-i and 3+\(\sqrt{2}\) are also roots of the given equation. Thus (x-(2+i)), (x-(2-i)), (x-(3-\(\sqrt{2}\))) and (x-(3+\(\sqrt{2}\))) are factors. Thus their product.
((x-(2+i))(x-(2-i))(x-(3-\(\sqrt{2}\)))(x-(3+\(\sqrt{2}\))) is a factor of the given polynomial equation.
That is, (x²-4x+5)(x²-6x+7) is a factor. Dividing the given polynomial equation by this factor, we get the other factor as (x²-3x-4) which implies that 4 and −1 are the other two roots. Thus
2+i, 2-i, 3+\(\sqrt{2}\), 3-\(\sqrt{2}\), -1, and 4 are the roots of the given polynomial equation.

Given equation is 2x⁴- 8x + 6x²- 3 = 0
Here a = 2, b = -8, c = 6, d = 0, e = -3
Let ∝, β, ૪ and \(\delta \) be the roots of equation (1)
Then by Vieta's formula,
\(\sum { _{ 1 }= } \alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -(-8) }{ 2 } =4\)
\(\sum { _{ 2 } } =\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { 6 }{ 2 } =3\)
\(\sum { _{ 3 } } =\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =\frac { 0 }{ a } \)
\(\sum { _{ 4 } } =\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { -3 }{ 2 } \)
Now, (a+b+c+d)² = a²+b²+c²+d²+2(ab+ac+ad+bc+cd)
⇒ n∝²+β²+૪²+\(\delta\)² = (∝ + β + ૪ + \(\delta\))²-2(\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta \))
∝² + β² + ૪² = 4²-2(3) = 16 - 6 = 10

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.

The roots of x³+2x²+3x+4 = 0 are ∝, β, ૪
∴ ∝ + β + ૪ = -co-efficient of x² = -2   .........(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3   ..........(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4  .......(3)
From the cubic equation whose roots are \(\frac { 1 }{ \alpha } ,\frac { 1 }{ \beta } ,\frac { 1 }{ \gamma } \)
\(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma } =\frac { \beta \gamma +\gamma \alpha +\alpha \beta }{ \alpha \beta \gamma } =\frac { 3 }{ -4 } =\frac { -3 }{ 4 } \)
\(\frac { 1 }{ \alpha \beta } +\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } =\frac { \gamma +\alpha +\beta }{ \alpha \beta \gamma } =\frac { -2 }{ -4 } =\frac { 1 }{ 2 } \)
\(\left( \frac { 1 }{ \alpha } \right) \left( \frac { 1 }{ \beta } \right) \left( \frac { 1 }{ \gamma } \right) =\frac { 1 }{ \alpha \beta \gamma } =\frac { 1 }{ -4 } =-\frac { 1 }{ 4 } \)
∴ The required cubic equation is 
\({ x }^{ 3 }-\left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma } \right) { x }^{ 2 }+\left( \frac { 1 }{ \alpha \beta } +\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } \right) x-\left( \frac { 1 }{ \alpha } ,\frac { 1 }{ \beta } ,\frac { 1 }{ \gamma } \right) \)
\(\Rightarrow { x }^{ 3 }+\frac { 3 }{ 4 } { x }^{ 2 }+\frac { 1 }{ 2 } x+\frac { 1 }{ 4 } =0\)
Multiplying by 4 we get, 
4x³ + 3x² + 2x + 1 = 0

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)
= \(\left( \frac { b }{ a } \right) ^{ 2 }-2\left( \frac { c }{ a } \right) \)
= \(\frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \).

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

\(x=\frac { c }{ a } \)

4) \({ ax }^{ 2 }+\frac { b }{ x } +c=0\)