Let f(x) = 8x³- 2x² - 7x +3 = 0
Here sum of the co-efficients of odd terms = 8 - 7 = 1
and sum of the co-efficients of even terms = -2 +3 = 1
Hence, x = - 1 is a root of f(x)
Let us divide f(x) by (x + 1)

∴ The other factor is 8x² - 10x + 3
\(\Rightarrow x=\frac { 10\pm \sqrt { 100-4(8)(3) } }{ 2\times 8 } \)
\(\Rightarrow x=\frac { 10\pm \sqrt { 100-96 } }{ 16 } \Rightarrow x=\frac { 10\pm 2 }{ 16 } \)
\(\Rightarrow x=\frac { 12 }{ 16 } \ or \)
\(x=\frac { 8 }{ 16 } \Rightarrow x=\frac { 3 }{ 4 } ,\frac { 1 }{ 2 } \)
∴ The roots are -1, \(\frac{1}{2},\frac{3}{4}\)

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.

Given cubic equation is 2x³-6x²+3x+k = 0
Here, a = 2, b = -6, c = 3, d = k
Let ∝, β, ૪ be the roots
Given ∝ = 2(β+૪) ⇒ \(\frac{\alpha}{2}\) = β+૪ ...(1)
Now, \(\alpha +\beta +\gamma =\frac { -b }{ a } =-\frac { (-6) }{ 2 } =3\)
\(\frac { \alpha }{ 2 } +\alpha =3\Rightarrow \frac { \alpha +2\alpha }{ 2 } =3\Rightarrow \frac { 3\alpha }{ 2 } =3\)
\(\Rightarrow \alpha =2\)
\(\alpha \beta \gamma =\frac { -d }{ a } =\frac { -k }{ 2 } \Rightarrow 2.\beta \gamma =\frac { -k }{ 2 } \)
\(\beta \gamma =\frac { -k }{ 4 } ...(2)\)
Also, \(\alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \) 
\(2\beta +\beta \gamma +2\gamma =\frac { 3 }{ 2 } \)
\(2(\beta +\gamma )+\beta \gamma =\frac { 3 }{ 2 } \)
\(\alpha \frac { -k }{ 4 } =\frac { 3 }{ 2 } \quad [from(1)\& (2)]\)
Also, \(2-\frac { k }{ 4 } =\frac { 3 }{ 2 } [\because \alpha =2]\)
\(2-\frac { 3 }{ 2 } =\frac { k }{ 4 } \Rightarrow \frac { 1 }{ 2 } =\frac { k }{ 4 } \)
\(\\ k=\frac { 4 }{ 2 } \Rightarrow k=2\)
From(2), \(\beta \gamma =\frac { -k }{ 4 } =\frac { -2 }{ 4 } =\frac { -1 }{ 2 } \)\(\Rightarrow \gamma =\frac { -1 }{ 2\beta } \)
From \((1),\beta +\gamma =\frac { \alpha }{ 2 } =\frac { 2 }{ 2 } =1\)
Substituting \(\gamma =\frac { -1 }{ 2\beta } \) We get
\(\beta -\frac { 1 }{ 2\beta } =1\Rightarrow 2{ \beta }^{ 2 }-1=2\beta \Rightarrow 2\beta -2\beta -1=0\)
\(\beta =\frac { 2\pm \sqrt { 4-4(2)(-1) } }{ 4 } =\frac { 2\pm \sqrt { 4+8 } }{ 4 } \)
\(=\frac { 2\pm \sqrt { 12 } }{ 4 } =\frac { 2\pm 2\sqrt { 3 } }{ 4 } \)
\(\beta =\frac { 1\pm \sqrt { 3 } }{ 2 } \)
Hence the roots are \(2,\frac { 1+\sqrt { 3 } }{ 2 } ,\frac { 1-\sqrt { 3 } }{ 2 } \) and k = 2

Let the roots form a GP be \(\frac{a}{\lambda}\), a, a\(\lambda\)
Product of the roots \(\frac{a}{\lambda} \times a \times a \lambda=\frac{24}{3}=8\)
a³ = 8
a = 2
Sum of the roots, \(\frac{a}{\lambda}+a+a \lambda=\frac{26}{3}\)
\( a\left(\frac{1}{\lambda}+1+\lambda\right) =\frac{26}{3} \)
\(2\left(\frac{1+\lambda+\lambda^{2}}{\lambda}\right) =\frac{26}{3} \)
\( 3+3 \lambda+3 \lambda^{2} =13 \lambda \)
\(3 \lambda^{2}+3 \lambda-13 \lambda+3 =0 \)
\(3 \lambda^{2}-10 \lambda+3 =0\)
\( (\lambda-3)(3 \lambda-1)=0 \)
\( \lambda=3 \text { (or) } \lambda=\frac{1}{3}\)
\( \lambda=3, \quad \frac{a}{\lambda}=\frac{2}{3} , \)
\( a \lambda=2(3)=6 \)
If \( \lambda=\frac{1}{3}, \frac{a}{\lambda}=\frac{2}{\frac{1}{3}}=6 \)
If \( a \lambda=2\left(\frac{1}{3}\right)=\frac{2}{3} \)
Roots are \(\frac{a}{\lambda},\ a,\ a \lambda\)
If \(\lambda\) = 3 roots are \(\frac{2}{3}\), 2, 6
If \(\lambda=\frac{1}{3}\) roots are 6, 2, \(\frac{2}{3}\)

Rearrange the terms as
(2x-1)(2x+ 3) (x + 3) (x - 2) + 20 = 0
⇒ (4x² + 6x - 2x- 3)(x² - 2x + 3x - 6) + 20 = 0
⇒ (4x² + 4x - 3) (x² + x - 6) + 20 = 0
put x²+ x = y
⇒ (4y - 3) (y - 6) + 20 = 0
⇒ 4y² - 24y - 3y + 18 + 20 = 0
⇒ 4y²-27y +38 = 0
⇒ (y - 2)( 4y - 19) = 0
\(y=2,\frac { 19 }{ 4 } \)

Case (i) 
When y = 2
x²+ x = 2
x² + x - 2 = 0
⇒ (x + 2)(x - 1) = 0
⇒ x = -2, 1
Case (ii) 
When \(y=\frac { 19 }{ 4 } ,{ x }^{ 2 }+x=\frac { 19 }{ 4 } \)
\(\Rightarrow { 4x }^{ 2 }+4x=19\)
\(\Rightarrow { 4x }^{ 2 }-4x-19=0\)
\(\Rightarrow x=\frac { -4\pm \sqrt { 16-4(4)(-19) } }{ 8 } \)
\(\Rightarrow x=\frac { -4\pm \sqrt { 16+304 } }{ 8 } \)
\(\Rightarrow x=\frac { -4\pm \sqrt { 320 } }{ 8 } \)
\(\Rightarrow x=\frac { 4\pm 8\sqrt { 5 } }{ 8 } \)
\(\Rightarrow x=\frac { -4(-1\pm 2\sqrt { 5 } ) }{ 8 } \)
\(\frac{-1 \pm 2 \sqrt{5}}{2}\)
Hence the roots are 1, -2, \(\frac{-1 \pm 2 \sqrt{5}}{2}\)

Let the roots be in G.P.
Then, we can assume them in the form \(\frac { \alpha }{ \lambda } \), α, αλ.
Applying the Vieta’s formula, we get
Σ₁ = \(\alpha \left( \frac { 1 }{ \lambda } +1+\lambda \right) =\frac { b }{ a } \) ....(1)
Σ₂ = \(\alpha ^{ 2 }\left( \frac { 1 }{ \lambda } +1+\lambda \right) =\frac { c }{ a } \) ............(2)
Σ₃ = α³ = -\(\frac { d }{ a } \) ..........(3)
Dividing (2) by (1), we get
α = -\(\frac { c }{ b } \) ...........(4)
Substituting (4) in (3), we get \(\left( -\frac { c }{ b } \right) ^{ 3 }=\frac { d }{ a } \) ⇒ ac³ = db³.

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.

The discriminant Δ =((4p)² - 4(4)(p+2) = 16(p²-p-2) = 16(p+1)(p-2). So we get
Δ < 0 if -1 < p < 2
Δ = 0 if p = -1 or p = 2
Δ >0 if \(\infty\). Thus the given polynomial has
imaginary roots if -1 < p < 2
equal real roots if p = −1 or p = 2;
distinct real roots if -\(\infty\) < p < -1 or 2 < p < \(\infty\)

Since two roots are in the ratio p : q : r, we can assume the roots as pλ, qλ and rλ .
Then, we get
Σ₁ = pλ + qλ + rλ = -a .....(1)
Σ₂ = (pλ)(qλ)+(qλ)(rλ)+(rλ)(pλ) ..........(2)
Σ₃ = (pλ)(qλ)(rλ) = -c .....(3)
Now, we get
(1) ⇒ λ = -\(\frac { a }{ p+q+r } \) ..........(4)
(3) ⇒ λ³ = \(\frac { c }{ pqr } \) ...........(5)
Substituting (4) in (5), we get
\(\left( \frac { a }{ p+q+r } \right) ^{ 3 }=-\frac { c }{ pqr } \) ⇒ pqra³ = c(p+q+ r)³.

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 \((\sqrt { 5 } -\sqrt { 3 } )\) 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.

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
\(\Rightarrow { x }^{ 2 }-x\left( \frac { 3 }{ 2 } \right) -10=0\)
\(\Rightarrow { 2x }^{ 2 }-3x-20=0\)

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