(2)

Let $p\left(x\right)=x^2-3x-m(m+3)$

$\Rightarrow p\left(x\right)=x^2-(m+3)x+mx-m(m+3)$

$=x\{x-(m+3\left)\right\}+m\{x-(m+3\left)\right\}$

For zeros of $p\left(x\right)$

$\Rightarrow p\left(x\right)=(x+m)\{x-(m+3\left)\right\}=0$$\Rightarrow x=-m,m+3$

$\therefore$ Its zeros are $-m,m+3$

(2)

Given, $f\left(x\right)=p{x}^2-2x+3p$

Since $\alpha$ and $\beta$ are the zeroes of the given polynomial.

$\therefore$ $\alpha +\beta =-\frac{(-2)}{p}=\frac{2}{p}, \text{and} \alpha \beta =\frac{3p}{p}=3$

$\because$ $\alpha +\beta =\alpha \beta$      (given)

$\therefore$ $\frac{2}{p}=3\Rightarrow p=\frac{2}{3}$

(1)

Given polynomials : $x^2+px+q$ ...(i)

and $4x^2-5x-6$       ...(ii)

Zero of polynomial $4x^2-5x-6$ are: $x=2$ and $x=-3/4$

Now, zero of polynomial $x^2+px+q$ are 4 and $-3/2$

$\therefore$ Sum of zeroes $=-\frac{p}{1}$

$\Rightarrow 4-\frac{3}{2}=-\frac{p}{1}\Rightarrow -p=\frac{5}{2}\Rightarrow p=-\frac{5}{2}$

(4)

The polynomials of the form ${(x+a)}^2$ and ${(x-a)}^2$ has only equal roots and graphs of these polynomials cut x-axis at only one point. These polynomials are quadratic Thus, Assertion is false Reason is true.

(2)

Sum of zeroes $=\sqrt{2}=\frac{-(-k\sqrt{2})}{2}\Rightarrow k=2$

(3)

$\alpha +\beta =\frac{-b}{a}=\frac{-3}{5}, \alpha \beta =\frac{c}{a}=\frac{-7}{5}$

Now, $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{\frac{-3}{5}}{\frac{-7}{5}}=\frac{3}{7}$

(1)     Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).