Let one root = \(\alpha \)
Other root = 6 \(\alpha \)
Sum of roots = \(\alpha \) + 6\(\alpha \) = \(\frac { 14 }{ k } \)
\(\Rightarrow 7\alpha =\frac { 14 }{ k } \)
Product of roots \(\alpha (6\alpha )=\frac { 8 }{ k } \)
\(\Rightarrow 6\alpha ^{ 2 }=\frac { 8 }{ k } \)
\(6\times \frac { 4 }{ k^{ 2 } } =\frac { 8 }{ k } \)
\(\Rightarrow \frac { 3 }{ k^{ 2 } } =\frac { 1 }{ k } \)
or 3k=k²
3k-k²
k[3-k]=0
k=0 or k=3
k=0 is not possible
Hence k=3

Putting x = -\(\frac{1}{2}\). in 3x² + 2kx - 3 = 0
\(3\left( -\frac { 1 }{ 2 } \right) ^{ 2 }+2k\left( -\frac { 1 }{ 2 } \right) -3=0\)
⇒ \(\frac{3}{4}\) - k- 3 = 0
⇒ k = \(\frac{3}{4}\)- 3
⇒ k = \(\frac{3-12}{4}\)
⇒ k = \(\frac{-9}{4}\)

Since x=2 is a root of given quadratic equation 
\(\therefore\) 3(2)²-P(2)-2=0
\(\Rightarrow\)  12-2p-2=0
\(\Rightarrow\)  p=5
Hence, the value of p is 5.

Given quadratic equation is
7x²-4a+3=0
Here a=7,b=-4,c=3
D=b²-4ac
= (-4)²-4x7x3
= 16-84=-68
\(\therefore \)D<0 therefore given quadratic equation has no real roots.

Given equation is \(4x^2+4\sqrt3x+3=0\)
Here a=4, b=4\(\sqrt3\), c=3
D=b²-4ac=48-48=0
As D=0, the equation has real and equal roots.

D=b²-4ac
D=(2k)²-4X3X27=4k²-324
For real and equal roots, D=0
4k²-324=0
4k²=324
k²=3244
k²=81
k=±9

Given equation is x²-9x+20=0
x²-5x-4x+20=0
x(x-5)-4(x-5)=0
(x-5)(x-4)=0
either x-5=0 or x-4=0
x=5 or x=4
x=4 or 5 are the roots/solution of the given quadratic equation

\(\sqrt{x^2+4}=(x^2+1)^2\)
\(=x^2+1+2x^2\)
\(\sqrt{x^2+4}-1=x^4+2x^2\)
\(x^2+4+1-2\sqrt{x^2+4}=(x^4+2x^2)^2\)
It is not a quadratic equation.

(x+1)(x+3)=(x-1)(x-4)
x²+4x+3=x²-5x+4
9x-1=0
Degree of equation is 1.
It is not a quadratic equation.

(2x+1)(x-3)=(x-1)²
2x²-6x+x-3=x²-2x+1
x²-3x-4=0
Which is of the form ax²+bx+c=0
Hence, given equation is a quadratic equation.

Given, equation is (x+2)³=2x(x²-1)
\(\Rightarrow\) x³ + 8 + 3x²(2) + 3x (2)² = 2x³ - 2x
[\(\because\) (a + b)³ = a³ + b³ + 3a²b + 3ab²]
\(\Rightarrow\) x³ + 8 + 6x² + 12x = 2x³ - 2x
\(\Rightarrow\) x³ + 8 + 6x² + 12x - 2x³ + 2x = 0
\(\Rightarrow\) -x³ + 6x² + 14x + 8 = 0
which is not of the form ax² + bx + c = 0,
because it has cubic term, i.e. x³.
\(\therefore\) It is not a quadratic equation.

\(\left[2\pm \sqrt {3}\right]\)

\(\left[{1\pm\sqrt {85}\over 6}\right]\)

Let total number of persons be x
\(\therefore \) Share of each person = Rs \(\frac { 9000 }{ x } \)
When the number of persons increased by 20
According to the question 
\(\frac { 9000 }{ x } -\frac { 9000 }{ x+20 } =160\Rightarrow 9000\left( \frac { 1 }{ x } -\frac { 1 }{ x+20 } \right) =160\)
\(\Rightarrow \frac { x+20-x }{ x(x+20) } =\frac { 160 }{ 9000 } \Rightarrow \frac { 5 }{ x^{ + }+20x } =\frac { 4 }{ 900 } =\frac { 1 }{ 255 } \)
\(\Rightarrow x^{ 2 }+20x=1125=\Rightarrow x^{ 2 }+20x-1125=0\)
\(\Rightarrow x^{ 2 }+45x-25x-1125=0\quad \Rightarrow x(x+45)-25(x+45)=0\)
\(\Rightarrow (x-25)(x+45)=0\Rightarrow x=25,45\) (rejected)
x=25
Original number of persons = 25

Let digit at unit's place = a and digit at ten's place = y
Number = 10y+x
A.T.Q xy=18 \(\Rightarrow y=\frac { 18 }{ x } \) ....(i)
and 10y+x-63=10x+y \(\Rightarrow 9y-9x-63=0\)
\(\Rightarrow y-x-7=0\Rightarrow \frac { 18 }{ x } -x-7=0\)[Using eq(i)]
\(\Rightarrow 18-x^{ 2 }-7x=0\Rightarrow x^{ 2 }+7x-18=0\) 
\(\Rightarrow (x+9)(x-2)=0\Rightarrow x=-9,x=2\)
When x=2 ,y =  \(\frac { 18 }{ 2 } =9\)
Number = 92

Let Varun's present age be x years and Swati's present age be y years
Case I : 7 years ago
Varun's age was (x-7) years and Swati's age was (y-7) years
ATQ   (x-7) = 5(y-7)²   \(\Rightarrow x=5(y-7)^{ 2 }+7\)  ...(i)
Case II : 3 years hence 
Varun's age will be (x+3) years and Swati's age will be (y+3) years 
A.T.Q \(y+3=\frac { 2 }{ 5 } (x+3)\)
\(\Rightarrow y+3=\frac { 2 }{ 5 } \left[ 5(y-7)^{ 2 }+7+3 \right] \) [Using eq.(i)]
\(\Rightarrow y+3=\frac { 2 }{ 5 } \times 5(y-7)^{ 2 }+\frac { 2 }{ 5 } \times 10\Rightarrow y+3=22(y^{ 2 }-14y+49)+4\)
\(\Rightarrow y+3=2y^{ 2 }-28y+98+4\Rightarrow 2y^{ 2 }-29y+99=0\)
\(\Rightarrow 2y^{ 2 }-18y-11y+99=0\Rightarrow 2y(y-9)-11(y-9)=0\)
\(\Rightarrow (y-9)(2y-11)=0\Rightarrow y=9,y=\frac { 11 }{ 2 } \) (rejecting)
\(\therefore y=9\)  \(\therefore x=5(9-7)^{ 2 }+7\) [From (i)]
x = 27
Present age of Swati = 9 years and Varun =27 years.

Let the two consecutive integers be x and x+1
ATQ x²+(x+1)2=365
\(\Rightarrow\) x²+x2+2x+1=365 \(\Rightarrow\) 2x²+2x-364=0
\(\Rightarrow\) x²+x-182=0 \(\Rightarrow\) x²+14x-13x-182=0
\(\Rightarrow\) x(x+14)-13(x+14)=0 \(\Rightarrow\) (x-13)(x+14)=0
\(\Rightarrow\) x=13, -14 (-14 is rejected because it is a negative integer)
Hence, the two consecutive positive integers are 13 and 13+1=14

Here x = - 2 is the root of the equation x² + k(4x + k -1) + P = 0
\(\Rightarrow 3(-2)^{ 2 }+7(-2)+p=0\)
\(\Rightarrow p=2\)
Root of the equation x² + 4kx + k² - k + 2 = 0 are equal
\(\Rightarrow 16k^{ 3 }-4(k^{ 2 }-k+2)=0\)
\(\Rightarrow 3k^{ 2 }+k-2=0\)
\(\Rightarrow (3k-2)(k+1)=0\)
\(\Rightarrow k=\frac { 2 }{ 3 } ,1\)

Let original duration of the flight be x hours.
 Distance = 2800 km 
\(\therefore \) Usual method = \(\frac { 2800 }{ x } km/h\)
When time = \(\left( x+\frac { 1 }{ 2 } \right) \) hrs
The new speed = \(\frac { 2800 }{ x+\frac { 1 }{ 2 } } =\frac { 5600 }{ 2x+1 } \)
ATQ \(\frac { 2800 }{ x } -\frac { 5600 }{ 2x+1 } =100\quad \Rightarrow \frac { 2800(2x+1)-5600x }{ (2x+1)x } =100\)
\(\Rightarrow 2800=100(2x^{ 2 }+x)\Rightarrow 2x^{ 2 }+x-28=0\)
\(\Rightarrow 2x^{ 2 }+8x-7x-28=0\quad \Rightarrow 2x(x+4)-7(x+4)=0\)
\(\Rightarrow (x+4)(2x-7)=0\quad \Rightarrow x=-4\) (rejected) or x=\(\frac { 7 }{ 2 } =3\frac { 1 }{ 2 } \)
\(\therefore \) Original duration = \(3\frac { 1 }{ 9 } \) hours