(a) A 2-digit number is seven times the sum of its digits and two (2) more than 5 times the product of its digits. Find the number.

Let digit at unit place be \(\displaystyle x\) and digit at tens place be \(\displaystyle y\)

\[\begin{aligned} & \therefore \text { number }=10 y+x \\ & \text { ATQ } \\ & \quad 10 y+x=7(x+y) \\ & \Rightarrow \quad 3 y=6 x \text { or } y=2 x \quad \ldots(1) \end{aligned} \]

Also, \(\displaystyle 10 \mathrm{y}+\mathrm{x}=5 \mathrm{xy}+2 \quad \ldots(2)\)

from ($\displaystyle 1$) and ($\displaystyle 2$), we get

\[\begin{aligned} & 10 x^{2}-21 x+2=0 \\ \Rightarrow \; & (x-2)(10 x-1)=0 \end{aligned} \]

\(\displaystyle \therefore \mathrm{x}=2\)

So, \(\displaystyle \mathrm{y}=4\)

∴ Required number is 42.

For real and equal roots, \(\displaystyle \mathrm{D}=0\)

\[\begin{aligned} & \therefore[-(\mathrm{p}+1)]^{2}-4(\mathrm{p}+4)=0 \\ & \Rightarrow \mathrm{p}^{2}-2 \mathrm{p}-15=0 \\ & \Rightarrow(\mathrm{p}-5)(\mathrm{p}+3)=0 \\ & \therefore \mathrm{p}=5,-3 \end{aligned} \]

For \(\displaystyle \mathrm{p}=5\),

\[\begin{aligned} & 9 x^{2}-6 x+1=0 \\ \Rightarrow & (3 x-1)(3 x-1)=0 \\ \therefore x= & \frac{1}{3}, \frac{1}{3} \end{aligned} \]

For \(\displaystyle \mathrm{p}=-3\),

\[\begin{aligned} & x^{2}+2 x+1=0 \\ \Rightarrow & (x+1)(x+1)=0 \\ \therefore x= & -1,-1 \end{aligned} \]

Hence roots are \(\displaystyle \frac{1}{3}, \frac{1}{3}\) and \(\displaystyle -1,-1\) for \(\displaystyle \mathrm{p}=5\) and \(\displaystyle \mathrm{p}=-3\) respectively.