(a) Find the value of 'p' for which the quadratic equation p (x-4)(x-2)+ (x-1)^2=0 has real and equal roots.

\[\begin{array}{l} p(x-4)(x-2)+(x-1)^{2}=0 \\ p\left(x^{2}-6 x+8\right)+x^{2}-2 x+1=0 \\ (p+1) x^{2}-(6 p+2) x+(8 p+1)=0 \\ a=p+1, b=6 p+2, c=8 p+1 \end{array} \]For real and equal roots, \[\begin{aligned} \therefore D= & 0 \Rightarrow b^{2}-4 a c=0 \\ \Rightarrow \quad & (6 p+2)^{2}-4(p+1)(8 p+1)=0 \\ & 36 p^{2}+24 p+4-4\left(8 p^{2}+9 p+1\right)=0 \\ & 4 p^{2}-12 p=0 \Rightarrow 4 p(p-3)=0 \\ \Rightarrow \quad & p=0,3 \end{aligned} \]Or Had Aarush scored $\displaystyle 8$ more marks in a Mathematics test, out of $\displaystyle 35$ marks, $\displaystyle 7$ times these marks would have been $\displaystyle 4$ less than square of his actual marks. How many marks did he get in the test?Let actual marks be $\displaystyle x$ \[\text { ATQ } \quad \begin{array}{l} 7(x+8)=x^{2}-4 \\ x^{2}-7 x-60=0 \\ x^{2}-12 x+5 x-60=0 \\ (x-12)(x+5)=0 \\ x=12, x=-5 \text { (rejecting) } \end{array} \] ∴ Actual marks obtained by Aarush = $\displaystyle 12$