\(2 x^{2}-3 x-1 \text { (v) } 4 x^{2}-x+1\)

\(x^{2}-(2+\sqrt{3}) x+(2-\sqrt{3})\)

The number of zeroes of p(x) are 4. (∵ the graph intersects the x-axis at four points)

Let the quadratic polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+B=4=-\frac{b}{a}\) 
\(\alpha \beta=1=\frac{c}{a}\) 
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x² – 4x +1.

\(-\frac{1}{4}, \frac{1}{4}\) 
Let the quadratic polynomial be ax² + bx + c, and its zeroes be α + ß
Given  \(\alpha+\beta=-\frac{1}{4}=-\frac{b}{a}\) 
\(\alpha \beta=\frac{1}{4}=\frac{c}{a}\) 
If a = 4, b = 1, c = 1
Therefore, the quadratic polynomial is 4x² + x +1.

Let the quadratic polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+\beta=-\frac{b}{a}=1\) 
\(\alpha \beta=\frac{c}{a}=1\) 
If a = 1, b = -1, and c = 1
Therefore, the quadratic polynomial is x² – x +1.

\(0, \sqrt{5}\) 
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+B=0=-\frac{b}{a}\) 
\(\alpha \beta=\sqrt{5}=\frac{c}{a}\) 
If \(a=1, b=0, c=\sqrt{5}\) 
Therefore, the quadratic polynomial is \(x^{2}+\sqrt{5}\)

P(x)=x²+ 7x+ 7
Here, a =1, b =7, c=7
∴ α, β are both zeroes of p(x)
\(∴\ \alpha+\beta={-b\over a}=-7\)
\(\alpha\beta={c\over a}=7\)
Now \({1\over \alpha}+{1\over \beta}-2\alpha\beta={\beta+\alpha\over \alpha\beta}-2\alpha\beta\)
\(={-7\over 7}-2\times7=-1-14=-15\)

\(α+β=-{b\over a}\)
⇒ α + β = -5 and αβ=\(c\over a\)⇒α, β=5
Now \(α^{-1}+ β^{-1}={1\over \alpha}+{1\over \beta}={-5\over 5}=-1\)

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 \(\pm 18\)

Since \(\alpha\) and \(\beta\) are the zeroes of the polynomial, then
\(\alpha +\beta =-\frac { Coefficient \ of\ x }{ Coefficient \ of \ { x }^{ 2 } } \)
\(\Rightarrow \quad \quad \alpha +\beta =-\left( \frac { -1 }{ 1 } \right) =1\)   .... (i)
Given \(\alpha-\beta=9\)  ... (ii)
From (i) and (ii), \(\alpha=5,\beta=-4\)
\(\alpha \beta =\frac { Constant \ term }{ Coefficient \ of \ { x }^{ 2 } } \)
\(\alpha \beta =-k\)
\(\Rightarrow \quad (5)(-4) = -k\)
\(\Rightarrow \quad k = 20\)

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

\({ x }^{ 2 }-4\sqrt { 3 } x+3=0\)
If \(\alpha\) and \(\beta\) are the zeroes of \({ x }^{ 2 }-4\sqrt { 3 } x+3\)
then \(\alpha+\beta=-\frac{b}{a}\)
\(\Rightarrow \quad \alpha +\beta =-\frac { \left( -4\sqrt { 3 } \right) }{ 1 } \)
\(\Rightarrow \quad \alpha +\beta=4\sqrt { 3 }\)
and \(\alpha\beta=\frac{c}{a}\)
\(\Rightarrow \quad \alpha\beta=\frac{3}{1}\)
\(\Rightarrow \quad \alpha\beta=3\)
\(\therefore \quad \alpha +\beta -\alpha \beta =4\sqrt { 3 } -3\)

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
\(=\left( \frac { 7 }{ 2 } \right) ^{ 2 }-3=\frac { 49 }{ 4 } -\frac { 3 }{ 1 } =\frac { 37 }{ 4 } \)

According to the question, sum of zeroes = 3
Product of zeroes = \(-\frac{2}{5}\)
The required quadratic polynomial
= x² - x (Sum of zeroes) + Product of zeroes
= x² - x(3) - \(\frac{2}{5}\)
= x² - 3x - \(\frac{2}{5}\)
= \(\frac{1}{5}\left( 5{ x }^{ 2 }-15x-2 \right) \)
\(\therefore\)  The quadratic polynomial is \(\left( 5{ x }^{ 2 }-15x-2 \right) \frac{1}{5}\)

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, \(x=\frac { 2 }{ \sqrt { 3 } } ,2\sqrt { 3 } \)

f(x) = x² - 2x
= x(x - 2)
i.e. f(x) = 0 \(\Rightarrow \)  x = 0 or x = 2
Hence zeroes are 0 & 2.

-7

k=6

x²-x-56

x²-6x+7

-2 and -3

α+β=7  ...(i)
α-β=1 ....(ii)
On solving Eqs. (i) and (ii), we get α=4 and β=3
Now, k=αβ=12

 Quadratic