The given quadratic polynomial is f(x) = x² + px + 45. Let \(\alpha\) and \(\beta\) be the zeroes of the given quadratic polynomial.
\(\therefore \quad \alpha +\beta =-p\) and \(\alpha\beta=45\)           ....(i)
Given, \(\left( \alpha -\beta \right) ^{ 2 }=144\)
\(\Rightarrow \quad \left( \alpha +\beta \right) ^{ 2 }-4\alpha \beta =144\)
\(\Rightarrow \quad \left( -p \right) ^{ 2 }-4\times 45=144\)
\(\Rightarrow \quad p^{ 2 }-180=144\)
\(\Rightarrow \quad p^{ 2 }=144+180=324\)
\(\therefore \quad p=\pm \sqrt { 324 } =\pm 18\)
Thus, the value of p is \(\pm 18\)

Let  p(x) = 2x² + 3x + \(\lambda \) 
One of the zero is \(\frac{1}{2}\),
so  \(p\left( \frac { 1 }{ 2 } \right) =0\)
\(\Rightarrow \quad 2\left( \frac { 1 }{ 2 } \right) ^{ 2 }+3\left( \frac { 1 }{ 2 } \right) +\lambda =0\)
\(\Rightarrow \quad \frac { 1 }{ 2 } +\frac { 3 }{ 2 } +\lambda =0\)
\(\Rightarrow \quad \lambda +2=0\)
\(\therefore \quad \lambda =-2\)
Now, Product of zeroes = \(\alpha \beta =\frac { constant\quad term }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(\Rightarrow \quad \alpha \left( \frac { 1 }{ 2 } \right) =\frac { \lambda }{ 2 } =-\frac { 2 }{ 2 } =-1\)
\(\Rightarrow \quad \alpha=-2\)
Hence, other zero = - 2

Given, f(x) = 14x² - 42k²x - 9
Let one zero be \(\alpha\)
\(\therefore\) The other = - \(\alpha\)
\(\therefore\)  Sum of zeroes = \(\alpha+(-\alpha)=0\)
Sum of zeroes \(=-\frac { Coefficient\quad of\quad x }{ Coefficient\quad of\quad { x }^{ 2 } } \)
According to the question,
Sum of zeroes = \(\frac { 42{ k }^{ 2 } }{ 14 } =3{ k }^{ 2 }\)
\(\therefore \quad 3{ k }^{ 2 }=0 \quad \Rightarrow \quad k = 0\)

\(\alpha+\beta=-\frac{b}{a}\)
\(=\frac { -\left( -6 \right) }{ 1 } =6\)
and \(\alpha\beta=\frac{c}{a}=\frac{k}{1}=k\)
\(\therefore \quad { \alpha }^{ 2 }+{ \beta }^{ 2 }=\left( \alpha +\beta \right) ^{ 2 }-2\alpha \beta =40\)
\(\Rightarrow \quad \left( 6 \right) ^{ 2 }-2k=40\)
\(\Rightarrow \quad 36-2k=40\)
\(\Rightarrow \quad 2k = - 4\)
\(\therefore \quad k = - 2\)

Since, - 1 is a zero of the polynomial
p(x) = kx² - 4x + k,
then p(-1) = 0
\(\therefore\)  k(-1)² - 4 (- 1) + k = 0
\(\Rightarrow\)  k + 4 + k = 0
\(\Rightarrow\)  2k + 4 = 0
\(\Rightarrow\)  2k = - 4
Hence, K = - 2

p(x) = ax² + bx + c
Let \(\alpha\) and \(\frac{1}\alpha\) be the zeroes of p(x),  then
Product of zeroes, \(\alpha\times\frac{1}\alpha=\frac{c}{a}\)
So,required condition is, c = a.

f(x) = 2x² - 7x + 3
Sum of roots = p + q \(=-\frac { Coefficient\quad of\quad x }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(=-\left( \frac { -7 }{ 2 } \right) =\frac { 7 }{ 2 } \)
Product of roots = pq \(=-\frac { Constant\quad term }{ Coefficient\quad of\quad { x }^{ 2 } } =\frac{3}{2}\)
We know that 
(p + q)² = p² + q² + 2pq
\(\Rightarrow\)  p² + q² = (p + q)² - 2pq
\(=\left( \frac { 7 }{ 2 } \right) ^{ 2 }-3=\frac { 49 }{ 4 } -\frac { 3 }{ 1 } =\frac { 37 }{ 4 } \)

Sum of zeroes = 6, Product of zeroes = 9
\(\therefore\)  Quadratic polynomial is x² - 6x + 9
Also x² - 6x + 9 = 0
\(\Rightarrow\)  (x - 3)(x - 3) = 0
\(\Rightarrow\)  x = 3, 3
Hence zeroes are 3, 3

\(p(x) = \sqrt { 3 } { x }^{ 2 }-8x+4\sqrt { 3 } \)
\(=\sqrt { 3 } { x }^{ 2 }-6x -2x+4\sqrt { 3 } \)
\(=\sqrt { 3 } x\left( x-2\sqrt { 3 } \right) -2\left( x-2\sqrt { 3 } \right) \)
\(=\left( \sqrt { 3 } x-2 \right) \left( x-2\sqrt { 3 } \right) \)
\(\therefore\)  Zeroes are, \(x=\frac { 2 }{ \sqrt { 3 } } ,2\sqrt { 3 } \)

k=6

\(k=\pm 3\)

\(-\frac { 145 }{ 12 } \)

\(-\frac { b }{ a } \) and \(\frac { c }{ b } \)

Let α and \(\frac { 1 }{ \alpha } \) be two zeroes of the given polynomial, which are reciprocal of each other.
On comparing the given polynomial with Ax²+Bx+C, we get
A=a²+9, B=13 and C=6a
Now, product of zeroes,
\(\alpha \times \frac { 1 }{ \alpha } =\frac { Constant \ term }{ Coefficient \ of \ x^{ 2 } } \) 
\(\\ \Rightarrow \ 1=\frac { 6\alpha }{ a^{ 2 }+9 }\) 
\( \\ \Rightarrow \ a^{ 2 }+9=6a\) 
\(\\ \Rightarrow \ a^{ 2 }-6a+9=0\) 
\(\\ \Rightarrow \ (a-3)^{ 2 }=0\ \ \left[ \because \quad (x-y)^{ 2 }=x^{ 2 }+y^{ 2 }-2xy \right] \) 
\(\\ \therefore \ a=3\)

\(a=-\frac { 3 }{ 2 } \)