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

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

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

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

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.

Here ∝ = 1, β = 1 and ૪ = -2
∴ The required cubic equation is
x³-(1+1-2)x²(1-2-2)x-(1)(1)(-2) = 0
x³- 0x²-3x+2 = 0
x³-3x-2 = 0

Let p(x) = x⁹ -5x² - 14x⁷ = 0
p(x) has only one sign change
Also p(-x) = (-x)⁹ - 5 (-x)⁸ - 14(-x)⁷ = 0
⇒ p(-x) = -x⁹ -5x⁸ + 14x⁷ = 0
p(-x) has only one sign change
∴ p(-x) has at most one positive and one negative root.

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 sum required quadrate 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 ∝, 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, \(\sum { \frac { \alpha }{ \beta \gamma } } =\frac { \alpha }{ \beta \gamma } +\frac { \beta }{ \gamma \alpha } +\frac { \gamma }{ \alpha \beta } \)
\(=\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } }{ \alpha \beta \gamma } \)
\(=\frac { { (\alpha +\beta +\gamma ) }^{ 2 }-2(\alpha \beta +\beta \gamma +\gamma \alpha ) }{ \alpha \beta \gamma } \)
\(=\frac { { \left( -\frac { b }{ a } \right) }^{ 2 }-2\left( \frac { c }{ a } \right) }{ -\frac { d }{ a } } \)
\(=\frac { \frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } }{ -\frac { d }{ a } } \Rightarrow \frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \times \frac { -a }{ d } \)
\(\therefore \frac { \sum { \alpha } }{ \beta \gamma } =-\frac { \left( { b }^{ 2 }-2ac \right) }{- ad } =\frac { 2ac-{ b }^{ 2 } }{ ad } \)

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)
Form a cubic equation whose roots are 2∝, 2β, 2૪
2∝+2β+2૪ = 2(∝+β+૪) = 2(-2) = -4 [from (1)]
4∝β+4β૪+4૪∝ = 4(∝β+β૪+୪∝) = 4(3) = 12 [from (2)]
(2∝)(2β)(2૪) = 8(∝β૪) = 8(-4) = -32 [from (3)]
∴ The required cubic equation is
x³-(2∝+2β+2૪)x² + (2∝β+2β૪+2୪∝)x - (2∝)(2β)(2૪) = 0
⇒ x³+(-4)x²+12x+32 = 0
⇒ x³+4x²+12x+32 = 0

The length and breadth of the cuboid are x + 1.
x + 2 and x + 3
[∵ they are increased by 1, 2,. 3 units]
Also volume = V + 52 .
[since V is increased by 52]
∴ V + 52 = (x +1)(x + 2)(x + 3) .........(1)
⇒ V = (x + 1) (x + 2) (x + 3) - 52
Here a = -1, β = -2, ૪ = -3
⇒ V = x³-x²(α+β+૪)+x(αβ+β૪+૪α)-αβ૪ = 52
⇒ V = x³-x²(-1-2-3)+x(2+6+3)-(-1)(-2)(-3) = 52
⇒ V = x³-x²(-6)+x(11)+6-52
⇒ x³ = x²+6x²+11x+6-52
⇒ 6x²+11x-46 = 0
⇒ (6x+23)(x-2) = 0
⇒ (6x+23)(x-2) = 0
⇒ x = 2

∴ Volume of the cube = x³ = 2³ = 8 c
Volume of acuboid = 52 + 8 = 60
[∵ x = \(\frac{-23}{6}\) is not possible as x represents the side of the cube]

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

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

Let P(x) be the polynomial under consideration.
The number of sign changes for P(x) and P(−x) are 2 and 1 respectively. Hence it has at most two positive roots and at most one negative root. Since the difference between number of sign changes in coefficients of P(−x) and the number of negative roots is even, we cannot have zero negative roots. So the number of negative roots is 1. Since the difference between number of sign changes in coefficient of P(x) and the number of positive roots must be even, we must have either zero or two positive roots. But as the sum of the coefficients is zero, 1 is a root. Thus we must have two and only two positive roots Obviously the other two roots are imaginary numbers.

x⁴+ 3x³- 3x - 1 = 0
Sum of the co-efficients = 1 + 3 - 3.- 1 = 0 
⇒x = 1 is a root ⇒(x - 1) is a factor

[Using synthetic division]
∴ x = 1, 1 are the roots and the remaining factor
\(\Rightarrow x=\frac { -3\pm \sqrt { 9-4(1)(1) } }{ 2 } \)
\(\Rightarrow x=\frac { -3\pm \sqrt { 5 } }{ 2 } \)
∴ The roots are 1, -1, \(\frac { -3+\sqrt { 5 } }{ 2 } ,\frac { -3-\sqrt { 5 } }{ 2 } \).

12x + 8x = 29x²- 4
This equation can be re-written as
12x³ - 29x² + 8x + 4 = 0

∴ x = 2 is a root and the remaining factor is 
12x²- 5x - 2
⇒ (3x-2)(4x+1) = 0
⇒ 3x-2 = 0 or 4x+1 = 0

⇒ x = \(\frac{2}{3}\)
x = \(\frac{-1}{4}\)
∴ The roots are 2, \(\frac{2}{3}\), \(\frac{-1}{4}\)

Since the sum of the co-efficients is
2 - 9 + 10 - 3 = 12 - 12 = 0
⇒ x = 1 is a root off (x)
∴ (x - 1) is a factor off (x)
To find the other factor, let us divide f(x) by x-1 

[Using synthetic division]
f(x) = (x-1)(x - 3)(2x - 1) = 0
⇒x - 1 = 0, x - 3 = 0 or 2x - 1 = 0
⇒ x = 1, x = 3x, x = \(\frac{1}{2}\)
Hence the roots are 1, 3, \(\frac{1}{2}\).

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, \(\frac{-1 \pm 2 \sqrt{5}}{2}\)

(i) 
Rearrange the terms as,
(x-5) (x+4) (x-7) (x+6) = 504
{-2, 3, -7, 8}
⇒ (x² -x - 20) (x² -x - 42) = 504
Put x²- x = y
⇒ (y-20)(y-42) = 504
⇒ y²-62y+840-504 = 0
⇒ y²-62y+336 = 0
⇒ (y - 56) (y - 6) = 0
⇒ y = 56, 6
Case (i)
When y = 56, x² - x = 56

⇒ x² - x - 56 = 0
⇒ (x - 8)(x + 7) = 0
⇒ x = 8, -7
Case (ii) 
When y = 6,
x²- x = 6
x²- x - 6 = 0
⇒ (x - 3)(x + 2) = 0
⇒ x = 3, - 2

Hence the roots are -2, 3, 8, -7

Since the coefficient of the equations 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.

Let ∝, β, ૪ be the roots of the equation 
Given \(\frac { \alpha }{ \beta } =\frac { 3 }{ 2 } \Rightarrow 2\alpha =3\beta \Rightarrow \alpha =\frac { 3 }{ 2 } \beta \)
\(\therefore \frac { 3 }{ 2 } \beta ,\beta ,\gamma \) are the roots of the given equation
Then by vieta's formula,
\(\frac { 3 }{ 2 } \beta +\beta +\gamma =\frac { -b }{ a } =\frac { -(-9) }{ 1 } =9\)
\(\frac { 5 }{ 2 } \beta +\gamma =9\Rightarrow \gamma =9-\frac { 5 }{ 2 } \beta \)
\(\Rightarrow \gamma =\frac { 18-5\beta }{ 2 } ...(2)\)
Also \(\frac { 3 }{ 2 } \beta (\beta )+\beta \gamma +\left( \frac { 3 }{ 2 } \beta \right) \gamma =\frac { c }{ a } =\frac { 14 }{ 1 } =14\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 5 }{ 2 } \beta \left( \frac { 18-5\beta }{ 2 } \right) =14\ [using\ (2)]\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 90\beta }{ 4 } -\frac { 25{ \beta }^{ 2 } }{ 4 } =14\)
Multiplying by \(4,6{ \beta }^{ 2 }+90{ \beta }-25{ \beta }^{ 2 }=56\)
\(19{ \beta }^{ 2 }-90{ \beta }+56=0\)
\(\Rightarrow ({ \beta }-4)(19{ \beta }-14)=0\)
\(\Rightarrow { \beta }=4\)
\({ \beta }=\frac { 14 }{ 19 } \)
When \(\beta\) = 4, the other roots are \(\frac { 3 }{ 2 } (4),4,\frac { 18-5 }{ 2 } (4)\)
\(\Rightarrow 6,4,-1\)

When \(\\ \beta =\frac { 14 }{ 19 } ,\) the other roots are \(\frac { 3 }{ 2 } \beta ,\beta \frac { 18-5\beta }{ 2 } [by(2)]\)
\(\Rightarrow \frac { 3 }{ 2 } \left( \frac { 14 }{ 19 } \right) ,\frac { 14 }{ 19 } ,\frac { 18-5\left( \frac { 14 }{ 19 } \right) }{ 2 } \Rightarrow \frac { 21 }{ 19 } ,\frac { 14 }{ 19 } ,\frac { 136 }{ 19 } \)

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 < \(\infty\)

Let  α, β and γ  be the roots of x³+ ax²+ bx + c = 0
Then, we get
Σ₁ = α + β + γ = -a ....(1)
Σ₂ = αβ + βγ + γα = b ...(2)
Σ₃ = αβγ = -c ...(3)
We have to form the equation whose roots are  α², β² and γ².
Using (1), (2) and (3), we find the following
Σ₁ = α² + β² + γ² = (α + β + γ )² - 2( αβ + βγ + γα) = (-a)² -2(b) = a²-2b,
Σ₂ = α²β² +  β²γ² + γ²α² = (αβ + βγ + γα)² - 2((αβ)( βγ)(γα) + (γα)(αβ))
= (αβ + βγ + γα)² - 2αβγ (β + γ + α) = (b)² - 2(-c)(-a) = b²-2ca
Σ₃ = α²β²γ² = (αβγ)² = (-c)² = c²
Hence, the required equation is
x³-(α² + β² + γ²)x² + (α²β² + β²γ² + γ²α²)x -  α²β²γ² = 0
That is, x³-(a²-2b)x² + (b²-2ca)x-c² = 0

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