Given quadratic equation is 7y² - 4y + 5 = 0
On comparing with ay² + by + c= 0, we get
a = 7, b = - 4 and c = 5
Now, D = b² - 4ac = (-4)² - 4(7) (5)
= 16 - 140 = -124 < 0
Since, D < 0, so given quadratic equation has no real roots.

\(x+\frac{1}{x}=3\). Multiplying throughout by x, we get
x² + 1 = 3x
i.e., x² - 3x + 1 = 0, which is a quadratic equation.
Here, a = 1, b = - 3, c = 1
So, b² - 4ac = 9 - 4 = 5 > 0
Therefore, \(x=\frac{3 \pm \sqrt{5}}{2}\)
So, the roots are \(\frac{3+\sqrt{5}}{2} \text { and } \frac{3-\sqrt{5}}{2}\)

\(\frac { 1 }{ (x-1)(x-2) } +\frac { 1 }{ (x-2)(x-3) } =\frac { 2 }{ 3 } \)
\(\Rightarrow \frac { x-3+x-1 }{ (x-1)(x-2)(x-3) } =\frac { 2 }{ 3 } \)
\(\Rightarrow \frac { 2x-4 }{ (x-1)(x-2)(x-3) } =\frac { 2 }{ 3 } \)
\(\Rightarrow \frac { 2(x-2) }{ (x-1)(x-2)(x-3) } =\frac { 2 }{ 3 } \)
\(\Rightarrow \frac { 2 }{ (x-1)(x-3) } =\frac { 2 }{ 3 } \)
\(\Rightarrow 3=(x-1)(x-3)\\ \Rightarrow x^{ 2 }-4x+3=3\\ \Rightarrow x^{ 2 }-4x=0\\ \Rightarrow x(x-4)=0\\ \Rightarrow x=0\quad or\quad x=4\)

\(3 x^{2}-2 \sqrt{6} x+2=3 x^{2}-\sqrt{6} x-\sqrt{6} x+2\)
\(=\sqrt{3} x(\sqrt{3} x-\sqrt{2})-\sqrt{2}(\sqrt{3} x-\sqrt{2})\)
\(=(\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})\)
So, the roots of the equation are the values of x for which
\((\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})=0\)
Now, \(\sqrt{3} x-\sqrt{2}=0 \text { for } x=\sqrt{\frac{2}{3}}\)
So, this root is repeated twice, one for each repeated factor \(\sqrt{3} x-\sqrt{2}\)
Therefore, the roots of 3x² - 2\(\sqrt6\)x + 2 = 0 are \(\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}\)

ATQ (60-t) = \(\frac { t^{ 2 } }{ 4 } -3\)
\(\Rightarrow 240-4t=t^{ 2 }-12\)
\(\Rightarrow t^{ 2 }+4t-252=0\)
\(\Rightarrow t^{ 2 }+18t-14t-252=0\)
\(\Rightarrow (t+18)(t-14)=0\)
\(\Rightarrow \) t=14,-18 [rejected]
\(\Rightarrow \) t=14 minutes

Kx²+1-2(k-41)x+x²=0
\(\Rightarrow kx^{ 2 }+x^{ 2 }-2(k-1)x+1=0\)
\(\Rightarrow (k+1)x^{ 2 }-2(k-1)x+1=0\)
Here,a=k +1 ,b=-2(k-1),c=1
D=b²-4ac
=[-2(k-1)²--4(k+1)(1)
=4(k-1)²-4(k+1)
Roots are equal
D=0
\(\Rightarrow 4(k-1)^{ 2 }-4(k+1)=0\)
\(\Rightarrow 4(k-1)^{ 2 }-(k+1)=0\)
\(\Rightarrow (k-1)^{ 2 }-k-1=0\)
\(\Rightarrow k^{ 2 }+1-2k-k-1=0\)
\(\Rightarrow k^{ 2 }-3k=0\)
\(\Rightarrow k(k-3)=0\)
k=0 or k-3=0 \(\Rightarrow k=3\)
Non zero value of k =3
after putting the value of k=3 in equation we get
3x3²+1-2(3-1)x+x² =0
\(\Rightarrow 3x^{ 2 }+1-2(2)x+x^{ 2 }=0\)
\(\Rightarrow 3x^{ 2 }+1-4x+x^{ 2 }=0\)
\(\Rightarrow 4x^{ 2 }-4x+1=0\)
\(\Rightarrow 4x^{ 2 }-2x-2x+1=0\)
\(\Rightarrow 2x(2x-1)-1(2x-1)=0\)
\(\Rightarrow (2x-1)\quad (2x-1)=0\)
2x-1=0 or 2x-1=0
2x-1 or 2x=1
\(x=\frac { 1 }{ 2 } \) or \(x=\frac { 1 }{ 2 } \)
Roots are \(x=\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \)

The equation has real roots if \(D\ge0\)
\((4)^2-4p\ge0\) 
i.e., \(16-4p\ge0 \ or \ 16\ge4p \ or \ p\le4\)

Let  \(\alpha \) and \(\beta \) re the roots of the equation x²-7x+2k=0
\(\Rightarrow \alpha +\beta =\left( \frac { -7 }{ 1 } \right) =7\)  ....(i)
and \(\alpha \beta =2k\)  ....(ii)
ATQ \(\alpha -\beta =1\Rightarrow (\alpha -\beta )^{ 2 }=1\)
\(\Rightarrow (\alpha +\beta )^{ 2 }=1\)   [using (a-b)² = (a+b)²-4ab]
\(\Rightarrow (7)^{ 2 }-4\times 2k=1\) [using (i) and (ii)]
\(\Rightarrow 49-8k=1\Rightarrow -8k=-48\Rightarrow k=6\)

Given quadratic equation is 2x²-6x+a=0
\(\alpha ,\beta \) are roots of the equation
\(\alpha +\beta =\frac { -b }{ a } \Rightarrow \alpha +\beta =-\frac { -6 }{ 2 } \Rightarrow \alpha +\beta =3\Rightarrow \alpha =3-\beta \)............... (i)
Also \(2\alpha +5\beta =12\)
\(\Rightarrow 2\left( 3-\beta \right) +5=12\)
\(\Rightarrow 6-2\beta +5\beta =12\Rightarrow 3\beta =6\Rightarrow \beta =2\)
where \(\beta =2\), eq.(i) becomes \(\alpha =3-2=1\)
Now,product of roots = \(\frac { c }{ a } \)
\(\Rightarrow \alpha .\beta =\frac { a }{ 2 } \Rightarrow 1\times 2=\frac { a }{ 2 } \Rightarrow a=4\)

Given equation is \(2x^{ 2 }+x+4=0\)
On dividing both sides by 2,we get
\(x^{ 2 }+\frac { 1 }{ 2 } x+2=0\Rightarrow x^{ 2 }+\frac { 1 }{ 2 } x=-2\)
On adding  \(\left[ \frac { 1 }{ 2 } coefficentofx \right] ^{ 2 }\)
i.e. \(\left[ \frac { 1 }{ 2 } x\frac { 1 }{ 2 } \right] ^{ 2 }=\frac { 1 }{ 16 } \) both sides,we get
\(x^{ 2 }+\frac { 1 }{ 2 } x+\frac { 1 }{ 16 } =-2+\frac { 1 }{ 16 } \)
\(\Rightarrow \left( x+\frac { 1 }{ 4 } \right) ^{ 2 }=-\left( \frac { \sqrt { 31 } }{ 4 } \right) ^{ 2 }\)
Which is not possible as the square of a real number cannot be negative. Therefore, the real roots of the equation \(2x^{ 2 }+x+4=0\) do not exist.

Given equations is \(x^{ 2 }-3x-10=0\)
\(\Rightarrow x^{ 2 }-5x+2x-10=0\)   
\(\left[ \because -5x(2)=-10\\ and-5+2=-3 \right] \)
\(\Rightarrow x(x-5)+2(x-5)=0\)
\(\Rightarrow x-5=0\quad \Rightarrow x=5\)
\(\Rightarrow x+2=0\Rightarrow x=-2\)
Hence, the roots of the equation 
\(x^{ 2 }-3x-10=0\quad are-2\quad and\quad 5\)