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

Since α, β, and γ are the roots of the equation x³+ px²+ qx + r = 0, we have
Σ₁ α + β + γ = -p and Σ₃ αβγ = -r
\(\Sigma \frac { 1 }{ \beta \gamma } =\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } +\frac { 1 }{ \alpha \beta } =\frac { \alpha +\beta +\gamma }{ \alpha \beta \gamma } =\frac { -p }{ -r } =\frac { p }{ r } \).

Since α and β are the roots of the quadratic equation, we have α + β =\(\frac { 7 }{ 2 } \) and αβ = \(\frac { 13 }{ 2 } \).
Thus, to construct a new quadratic equation,
Sum of the roots = α²+β² = (α+β)²-2αβ =\(\frac { -3 }{ 4 } \)
Product of the roots = α²β² = (αβ)² = \(\frac { 169 }{ 4 }\)
Thus a required quadratic equation is  x²+\(\frac { 3 }{ 4 } x+\frac { 169 }{ 4 } \)= 0. 
From this we see that 4x²+3x+169 = 0 is a quadratic equation with roots α² and β².

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.

(x-4)(x-7)(x-2)(x+1) = 16
Given (x - 4)(x -7)(x - 2)(x + 1) = 16
Rearrange the terms as
(x - 4)(x - 2)(x -7)(x + 1) = 16
⇒ (x² - 6x + 8)(x2 - 6x -7) = 16
Put x²-6x = y
⇒ (y + 8) (y - 7) = 16
⇒ y² +y - 56 - 16 = 0
⇒ (y+9)(y-8) = 0
⇒ y = -9, 8

case (i)
 When y = -9
⇒ x²-6x = -9
⇒ x²-6x+9 = 0
⇒ (x-3)² = 0
⇒ x = 3, 3

case ii
When y = 8,
\(\Rightarrow x=\frac { 6\pm \sqrt { 36-4(1)(-8) } }{ 2 } \)
\(\Rightarrow x=\frac { 6\pm \sqrt { 36-32 } }{ 2 } \Rightarrow x=\frac { 6\pm \sqrt { 68 } }{ 2 } \)
\(\Rightarrow x=\frac { 6\pm 2\sqrt { 17 } }{ 2 } \Rightarrow x=\frac { 2(3\pm \sqrt { 17 } ) }{ 2 } \)
\(\Rightarrow x=3\pm \sqrt { 17 } \)
∴ The roots are 3, 3, 3 +\(\sqrt{17}\), and 3 -\(\sqrt{17}\)

Put x² = y
⇒ y²- 14y + 45 = 0
⇒ (y - 9) (y - 5) = 0
⇒ y = 9 or y = 5
⇒ x² = 9 or x² = 5
⇒ x = ±33 or x =±3√5
Hence the roots are 3, -3, √5 and -√5.

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 \(\frac { 5\pm \sqrt { 77 } }{ 2 } ,\frac { 5\pm \sqrt { 5 } }{ 2 } \).

The given equation is
x⁴- 9x² + 20 = 0
This is a fourth degree equation. If we replace x² by y then we get the quadratic equation
y²- 9y + 20 = 0
It is easy to see that 4 and 5 as solutions for y²- 9y + 20 = 0. Now taking x² = 4 and x² = 5, we get 2, -2, \(\sqrt{5}\), -\(\sqrt{5}\) as solutions of the given equation.
We note that the technique adopted above can be applied to polynomial equations like x⁶-17x³+30 = 0, ax^2k+ bx^k + c = 0 and in general polynomial equations of the form a_nx^kn + a_n-1x^k(n-1) + .... + a₁x^k + a₀ = 0 where k is any positive integer.

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

Given cubic equation 3x²-16x²+23x-6 = 0
Let ∝, \(\frac{1}{\alpha}\) and ૪ be the roots of the equation
[∵ product of two roots is 1]
\((1)\rightarrow { x }^{ 2 }-\frac { 16 }{ 3 } { x }^{ 2 }+\frac { 23 }{ 3 } -2=0 \) ......(1)
comparing  (1) with 
\({ x }^{ 3 }-\left( \frac { \alpha +\beta +\gamma }{ \alpha } \right) +\left( \alpha \frac { 1 }{ \alpha } +\frac { 1 }{ \alpha } .\gamma +\gamma \alpha \right) \)
\(-\alpha \frac { 1 }{ \alpha } .\gamma =0\) ..........(2)
We get, 
\(\alpha +\frac { 1 }{ \alpha } +\gamma =\frac { 16 }{ 3 } \) .......(3)
\(1+\frac { \gamma }{ \alpha } +\gamma \alpha =\frac { 23 }{ 3 } \)
\(\alpha .\frac { 1 }{ \alpha } .\gamma =2\Rightarrow \gamma =2\) ............(4)
Substituting ૪ = 2 in (3)
 \(\alpha +\frac { 1 }{ \alpha } +2=\frac { 16 }{ 3 } \)
\(\Rightarrow \alpha +\frac { 1 }{ \alpha } =\frac { 16 }{ 3 } -2=\frac { 16-6 }{ 3 } =\frac { 10 }{ 3 } \)
\(\Rightarrow \frac { { \alpha }^{ 2 }+1 }{ \alpha } =\frac { 10 }{ 3 } \)

\({ 3x }^{ 2 }+3=10\alpha \)
\({ 3\alpha }^{ 2 }-10\alpha +3=0\)
\(\alpha =\frac { -10 }{ 3 } or\ \alpha =\frac { 1 }{ 3 } \)
\((3\alpha +10)(3\alpha -1)=0\)
\(\alpha =\frac { -10 }{ 3 } \) is not possible \(\Rightarrow \alpha =\frac { 1 }{ 3 } \)
[\(\because \alpha =\frac { -10 }{ 3 } \) will not satisfy(5)]
∴ The roots are 3, \(\frac{1}{3}\), 2.

If P(x) denotes the polynomial in the equation, then P(2) = 0.
Hence 2 is a root of the polynomial.
To find other roots, we divide the given polynomial x³−5x²−4x + 20 by x − 2 and get Q(x) = x² −3x−10 as the quotient.
Solving Q(x) = 0 we get −2 and 5 as roots.
Thus 2, −2, 5 are the solutions of the given equation.

Since 2-\(\sqrt{3}\)i is a root of the required polynomial equation with real coefficients, 2 +\(\sqrt{3}\)i is also a root. Hence the sum of the roots is 4 and the product of the roots is 7. Thus x²-4x + 7= 0 is the required monic polynomial equation.

Given ∝, β and ૪ are the roots of ax³ + bx² + cx + d = 0
\(\therefore \alpha +\beta +\gamma =\frac { -b }{ a } \)
\(\alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \)
\(\alpha \beta \gamma =\frac { -d }{ a } \)
Now, \(\sum { \frac { \alpha }{ \beta \gamma } } =\frac { \alpha }{ \beta \gamma } +\frac { \beta }{ \gamma \alpha } +\frac { \gamma }{ \alpha \beta } \)
\(=\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } }{ \alpha \beta \gamma } \)
\(=\frac { { (\alpha +\beta +\gamma ) }^{ 2 }-2(\alpha \beta +\beta \gamma +\gamma \alpha ) }{ \alpha \beta \gamma } \)
\(=\frac { { \left( -\frac { b }{ a } \right) }^{ 2 }-2\left( \frac { c }{ a } \right) }{ -\frac { d }{ a } } \)
\(=\frac { \frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } }{ -\frac { d }{ a } } \Rightarrow \frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \times \frac { -a }{ d } \)
\(\therefore \frac { \sum { \alpha } }{ \beta \gamma } =-\frac { \left( { b }^{ 2 }-2ac \right) }{- ad } =\frac { 2ac-{ b }^{ 2 } }{ ad } \)