Let p(x) = x⁹ + 9x⁷ + 7x⁵ + 5x³ + 3x
p(x) has no sign change.
p(-x)⇒ (-x)⁹ + 9(-x)⁷ + 7(-x)⁵ + 5(-x)³ + 3(-x)
= -x⁹ - 9x⁷ - 7x⁵- 5x³ -3x
p(-x) also has no sign change
∴ p(x) has no positive and no negative root.

The roots are rational if Δ = b²−4ac = (−2p)²−4(p²−q²+2qr−r²).
But this expression reduces to 4(q²−2qr+r²) or 4(q−r)² which is a perfect square.
Hence the roots are rational.

Given polynomial equation is 
2x⁴+ 5x³−7x² + 8 = 0
Here a = 2, b = 5, c = -7, d = 0, e = 8
By Vieta's formula, 
\(\alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -5 }{ 2 } \)
\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { -7 }{ 2 } \)
\(\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =0\)
\(\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { 8 }{ 2 } =4\)
Given roots of quadratic equation are 
∝ + β + ૪ + \(\delta \) and ∝β૪\(\delta \)
∴ sum of the roots = (∝+β+૪+\(\delta \)) (∝β૪\(\delta \))
\(=\left( \frac { -5 }{ 2 } +4 \right) =\frac { -5+8 }{ 2 } =\frac { 3 }{ 2 } \)
\(=\left( \alpha +\beta +\gamma +\delta \right) (\alpha \beta \gamma \delta )\)
\(=\left( \frac { -5 }{ 2 } \right) (4)=\frac { -20 }{ 2 } =-10\)
∴ The required quadratic equation is x²-x 
(sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x\left( \frac { 3 }{ 2 } \right) -10=0\)
\(\Rightarrow { 2x }^{ 2 }-3x-20=0\)

Given roots are 1, 2 and 3
Here a = 1, β = 2 and ૪ = 3
A cubic polynomial equation whose roots are α, β, ૪ is
x³-(α+β+૪)+x²(αβ+β૪+૪α)x-∝β૪ = 0
⇒ x³-(1+1+2)x²(2+6+3)x-6 = 0
⇒ x³-6x²+11x-6 = 0

Let p(x) = x⁷-6x⁶+7x⁵+5x²+2x+2
It has only one change of sign. Now,
p(-x) = (-x)⁷ -6(-x)⁶ +7(-x)⁵ +5(-x)² +2(-x)+2
= -x⁷ -6x⁶ -7x⁵ + 5x² - 2x + 2
It has two, change of sign.
Hence, p(x) has one positive root and has at least two negative roots.

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

The quadratic equation 3x² + 2(a² + 1)x + (a² - 3a + 2)
Will have two roots of opposite sign if it has real roots and the product of the roots is negative.
⇒ 4(a²+1)²-12(a²-3a+2)\(\ge\) 0 and \(\frac { { a }^{ 2 }-3a+2 }{ 3 } <0\)
Both of these conditions are true if 
⇒ a²-3a+2< 0
⇒ (a-1)(a-2)< 0
⇒ 1<a<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

Sum of the roots = sin ∝ + cos ∝ = \(\frac{-b}{a}\)
Product of the roots = sin ∝ cos ∝ = \(\frac{c}{a}\)
Now 1 = cos²∝ + sin² ∝
= (sin ∝ +cos ∝)² - 2 sin ∝ cos ∝
\(1=\frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } \Rightarrow 1=\frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \)
⇒ a² = b² - 2ac ⇒ a² + 2ac = b²
Adding c² both sides, a² +2ac+c² = b²+c²
⇒ (a+c)² = b² + c²

Here ∝ = 2, β = -2 and ૪ = 4
x³-(2-2+4)x²+(- 4 - 8 + 8x)x-(2)(-2)(4) = 0
⇒x³- 4x²- 4x+16 = 0

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

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.

6x⁴- 35x³+ 62x²- 35x + 6 = 0
This equation is type I even degree reciprocal equation.
Hence, it can be rewritten as
\(6\left( { x }^{ 2 }+\frac { 1 }{ x } \right) -35(x+\frac { 1 }{ x } )+62=0 ...(1)\)
putting \(x+\frac { 1 }{ x } =y\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2={ y }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ={ y }^{ 2 }-2\)
∴ (1) becomes as,
\(\Rightarrow 6({ y }^{ 2 }-2)-35y+62=0\)
\(\Rightarrow { 6y }^{ 2 }-12-35y+62=0\)
\(\Rightarrow { 6y }^{ 2 }-35y+50=0\)
\(\Rightarrow (3y-10)(2y-5)=0\)
\(\Rightarrow y=\frac { 10 }{ 3 } ,\frac { 5 }{ 2 } \)
Case (i) when \(y=\frac { 10 }{ 3 } ,x+\frac { 1 }{ x } =\frac { 10 }{ 3 } \)


\(\Rightarrow \frac { { x }^{ 2 }+1 }{ x } =\frac { 10 }{ 3 } \)
\(\Rightarrow { 3x }^{ 2 }-10x+3=10x\)
\(\Rightarrow { 3x }^{ 2 }-10x+3=0\)
\(\Rightarrow (x-3)(3x-1)=0\)
\(\Rightarrow x=3,\frac { 1 }{ 3 } \)
Case (ii) when \(y=\frac { 5 }{ 2 } ,x+\frac { 1 }{ x } =\frac { 5 }{ 2 } \Rightarrow \frac { { x }^{ 2 }+1 }{ x } =\frac { 5 }{ 2 } \)
\(\Rightarrow { 2x }^{ 2 }+2=5x\Rightarrow { 2x }^{ 2 }-5x+2=0\)
\(\Rightarrow (x-2)(2x-1)=0\)
Hence the roots are \(2,\frac { 1 }{ 2 } ,3,\frac { 1 }{ 3 } \)

Let P(x) be the polynomial under consideration.
The number of sign changes for P(x) and P(−x) are zero and hence it has no positive roots and no negative roots. Clearly zero is not a root. Thus the polynomial has no real roots and hence all roots of the polynomial are imaginary roots.

Let the roots be in A.P. Then, we can assume them in the form α-d, α, α+d
Applying the Vieta’s formula (α-d)+α+(α+d) = \(\frac{p}{1}\) = p ⇒ 3α = -p ⇒ α = -\(\frac{p}{3}\).
But, we note that α is a root of the given equation. Therefore, we get
\(\left( -\frac { p }{ 3 } \right) ^{ 3 }+p\left(- \frac { p }{ 3 } \right) ^{ 3 }+q\left(- \frac { p }{ 3 } \right) ^{ 3 }+r\) = 0 ⇒ 9 pq = 2p³+27r.