Discriminant of the given quadratic equation is,
$$D =  4\left( a+1 \right) ^{ 2

}-4\left( a+1 \right) \left( a-2 \right) =4\left( a+1 \right) \left(

a+1-a+2 \right) =12\left( a+1 \right) $$
Now for two distinct roots,
$$D>0\Rightarrow a+1>0$$
$$\Rightarrow a \in (-1,\infty)$$. 

Given equation is,

$$x^2 + 4x + 1 = 0 $$
$$D = b^2 - 4ac$$
$$D = (4)^2 - 4(1)(1)$$
$$D = 16 - 4$$
$$D = 12 $$
$$D^2 = {12}^2 = 144$$

For the quadratic equation, $$(m\, -\, 1)\, x^{2}\, -\, 2\, (m\, -\, 1)\, x\, +\, 1\, =\, 0$$ 
Discriminant = $${(-2(m-1))}^2 - 4\times(m-1)= 0$$...... Since roots are real equal roots
$$4m^2 -8m + 4 - 4m + 4 = 0$$
$$4m^2 - 12m + 8 =0 $$
$$m^2 - 3m + 2 = 0$$
$$\therefore m = 1, 2 $$
Neglect $$m= 1$$ as the equation will not be quadratic anymore.
Hence, $$m = 2$$

Since, the roots are real and equal, $$D = 0$$
$$D=\sqrt {b^2-4ac=0}$$

Given equation is:

Given equation is $$2x^{2}\, +\, \displaystyle \frac{x\, -\, 1}{5}\, =\, 0$$
$$\therefore 10x^2 + x - 1 = 0$$
Applying qudratic formula,
$$x = \displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$a=10, b=1, c=-1$$

We have,
$$4{ x }^{ 2 } - \sqrt { 3 } x - 5 = 0$$
Divide whole equation by $$4$$,
$$\therefore   { x }^{ 2 } - \displaystyle\frac { \sqrt { 3 }  }{ 4 } x - \displaystyle\frac { 5 }{ 4 } = 0$$
$$\therefore { \left( x -\displaystyle\frac { \sqrt { 3 }  }{ 8 }  \right)  }^{ 2 } - \displaystyle\frac { 5 }{ 4 } = \displaystyle\frac { 3 }{ 64 }$$
$$\Rightarrow 4{ \left( x -\displaystyle\frac { \sqrt { 3 }  }{ 8 }  \right)  }^{ 2 } - 5 - \displaystyle\frac { 3 }{ 16 } = 0$$
$$\therefore \displaystyle \frac { 3 }{ 16 }$$ must be added to make $$4{ x }^{ 2 } - \sqrt { 3 } x - 5 = 0$$ a complete square

If $$b^{2}-4ac\geq 0$$
Then the roots are real.
Hence
By using quadratic formula, we get the roots as
$$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$
Hence, option 'D' is correct.

$$3p^2+4=-7p$$                      

Given, $$ 2y^2+5y+1=0$$
Compare the given equation with  $$ax^2+bx+c=0$$
Since, $$x=\frac {-b\pm\sqrt {b^2-4ac}}{2a}$$
In given quadratic equation  $$a=2 , b=5 , c=1$$
$$\therefore  y=\frac {-5\pm\sqrt {(5)^2-4\times 2 \times 1}}{2\times 2}$$
$$\therefore  y=\frac {-5\pm\sqrt {17}}{4}$$
equation is false.
Answer is 0.

$$\displaystyle B^{2}-4AC> 0$$ for 1st equation. 
$$\displaystyle \therefore 4\left ( c^{2}-ab \right )> 0 \Rightarrow c^{2}-ab$$ is + ive.$$\displaystyle B^{2}-4AC$$
for the 2nd equation is $$\displaystyle 4\left ( a+b \right )^{2}-4\left ( a^{2}+b^{2}+2c^{2} \right )$$ 
$$\displaystyle =4\left [ 2ab-2c^{2} \right ]=-8\left ( c^{2}-ab \right )=-ive.$$ as $$\displaystyle =c^{2}-ab$$ is+ive.
Hence the roots of the second equation are imaginary.

Let $$\displaystyle { \alpha  }_{ 1 },{ \beta  }_{ 1 }$$ be the roots of $$a{ x }^{ 2 }+bx+c=0$$ and $${ \alpha  }_{ 2 },{ \beta  }_{ 2 }$$ be the roots of $$p{ x }^{ 2 }+qx+r=0$$. Then

$$\displaystyle x^{2}-5x+6=0$$

Dividing the entire equation by 2, we get
$$x^{2}-\dfrac{5x}{2}+\dfrac{3}{2}=0$$

$$(x-\dfrac{5}{4})^{2}-\dfrac{25}{16}+\dfrac{24}{16}=0$$

$$(x-\dfrac{5}{4})^{2}=\dfrac{1}{16}$$

$$x-\dfrac{5}{4}=\pm\dfrac{1}{4}$$

$$x=\dfrac{5\pm1}{4}$$
Hence
$$x=\dfrac{3}{2}$$ and $$x=1$$

$$3x^{2}-2\sqrt{6}x+2=0$$
Dividing the entire equation, by $$ 3$$, we get
$$x^{2}-\dfrac{2\sqrt{6}}{3}+\dfrac{2}{3}=0$$
$$\left (x-\dfrac{\sqrt{6}}{3}\right)^{2}-\dfrac{6}{9}\dfrac{2}{3}=0$$
$$\left (x-\dfrac{\sqrt{6}}{3}\right)^{2}=0$$
$$x=\dfrac{\sqrt{6}}{3}$$
Hence, no rational roots.

Given quadratic equation is $$\sqrt 7 y^2 - 6y - 13 \sqrt 7 = 0 $$   

Comparing it with the standard form of equation $$ax^2+bx+c$$ we get $$a=7,b=6,c=13\sqrt 7$$ 
Therefore, $$x= \dfrac { -b\pm \sqrt { b^ 2-4ac }  }{ 2a } $$
$$=   \dfrac {  -(-6)\pm \sqrt { (-6)^{ 2 }-4\times \sqrt { 7 } \times \left( -13\sqrt { 7 }  \right)  }  }{ 2\sqrt { 7 }  } $$
$$=\dfrac {  6\pm \sqrt { 36+364 }  }{ 2\sqrt { 7 }  } $$
$$= \dfrac {  6\pm \sqrt { 400 }  }{ 2\sqrt { 7 }  } $$
$$=\dfrac {  6\pm 20 }{ 2\sqrt { 7 }  } $$
$$=\dfrac {  6- 20 }{ 2\sqrt { 7 }  } $$
$$=\dfrac { \\- 14 }{ 2\sqrt { 7 }  } $$
$$=\dfrac { \\ -7 }{ \sqrt { 7 }  } $$
$$=\dfrac { \\ -7\times \sqrt { 7 }  }{ \sqrt { 7 } \times \sqrt { 7 }  } $$
$$=-\sqrt 7$$
$$=\dfrac {  6+ 20 }{ 2\sqrt { 7 }  } $$
$$=\dfrac { \\ 26 }{ 2\sqrt { 7 }  } $$
$$=\dfrac { \\ 13 }{ \sqrt { 7 }  } $$

Given, $$x^2 - 6x + 4 = 0$$
$$\Rightarrow x^2-2\times x\times 3+3^2-5=0$$
$$\Rightarrow (x-3)^2=5$$
$$\Rightarrow x-3=\pm \sqrt 5$$
$$\Rightarrow x=3\pm \sqrt 5$$

The quadratic equation is $$2x^2 - 2\sqrt 2x + 1 = 0$$.
Comparing it with standard form of quadratic equation $$ax^2+bx+c=0$$,we get $$a=2,b=- 2\sqrt 2,c=1$$ Therefore the roots of the quadratic equation are,
$$x= \dfrac { -b\pm \sqrt { b^{ 2 }-4ac }  }{ 2a } $$
  $$= \dfrac { -\left( -2\sqrt { 2 }  \right) \pm \sqrt { \left( -2\sqrt { 2 }  \right) ^{ 2 }-4\times 2\times 1 }  }{ 2\times 2 } $$
  $$=\dfrac { \left( 2\sqrt { 2 }  \right) \pm \sqrt { 8-8 }  }{ 4 } $$
  $$=\dfrac { 2\sqrt { 2 } \pm 0 }{ 4 } $$
  $$=\dfrac { 2\sqrt { 2 }  }{ 4 } $$  
  $$=\dfrac { \sqrt { 2 }  }{ 2 } $$ 
  $$=\dfrac{ 1 }{ \sqrt { 2 }  } $$ 

Nature of the roots for a quadratic equation $$ax^2+bx+c=0$$ , can be determined by its discriminant

$$\Rightarrow x^{2}-2(8)x+8^{2}=0$$
This is the case of $$(x-a)^{2}=x^{2}-2ax+a^{2}$$
Therefore, $$(x-8)^{2}=0$$
$$\Rightarrow x=8,8$$

Given, $$\dfrac{x}{x - 1} + \dfrac{x - 1}{x} = 4$$
$$\Rightarrow \dfrac { x^{ 2 }+(x-1)^{ 2 } }{ x(x-1) } =4$$
$$\Rightarrow x^2+x^2-2x+1=4x^2-4x$$
$$\Rightarrow 2x^2-4x^2-2x+4x+1=0$$
$$\Rightarrow -2x^2+2x+1=0$$
$$\Rightarrow 2x^2-2x-1=0$$
Comparing $$2x^2-2x-1=0$$ with $$ax^2+bx+c=0$$, we get $$a=2, b=-2. c=-1$$
Now the roots are $$=$$ $$\dfrac { -b \pm \sqrt { b^{ 2 }-4ac }  }{ 2a } $$
$$=$$ $$\dfrac { -(-2)\pm \sqrt { (-2)^ 2-4\times 2\times (-1) }  }{ 2\times 1 }$$
$$=$$ $$ \dfrac { 2\pm \sqrt { 4+8 }  }{ 2 }$$
$$=$$ $$\dfrac { 2\pm \sqrt { 12 }  }{ 2 }$$
$$=$$ $$\dfrac { 2\pm 2\sqrt { 3 }  }{ 2 }$$
$$=$$ $$1\pm \sqrt 3$$

Given, $$(x^{2}-2x)^{2}-4(x^{2}-2x)+3=0$$

Given, $$x+\dfrac{1}{x}=3$$
$$\Rightarrow \dfrac{x+1}{x}=3$$
$$\Rightarrow x^2-3x+1=0$$
Comparing $$x^2-3x+1=0$$ with $$ax^2+bx+c=0$$, we get $$a=1, b=3. c=14$$
Now the roots are $$=$$ $$\dfrac { -b \pm \sqrt { b^{ 2 }-4ac }  }{ 2a } $$
$$=$$ $$\dfrac { -(-3)\pm \sqrt { (3)^ 2-4\times 1\times 1 }  }{ 2\times 1 } $$
$$=$$ $$\dfrac { 3\pm \sqrt { 9-4 }  }{ 2 }$$
$$=$$ $$ \dfrac { 3\pm \sqrt { 5 }  }{ 2 }$$
$$=$$ $$ \dfrac { 3+ \sqrt { 5 }  }{ 2 }$$ or $$\dfrac { 3- \sqrt { 5 }  }{ 2 }$$