\(x^{2}+a x+c=0 \)
\(x^{2}+d x+b=0 \)
\(8 \& 2 \text { are the roots }\)\(\text { 3. } 3 \text { are the roots }\)
\(\therefore a=-10 ; c=16 \quad d=-6, \quad b=9\)
\(x^{2}+a x+b =0 \)
\(x^{2}-10 x+9 =0 \)
\(\Rightarrow(x-1)(x-9) =0 \)
\(\therefore x =1 \text { (or) } 9 \)

\(3 x^{2}-5 x-7=0\)
\(\text { Let the roots be } \alpha, \beta\)
\(\Rightarrow \operatorname{Sum}: \alpha+\beta=\frac{5}{3}\)
\(\text { Product : } \alpha \beta=\frac{-7}{3}\)
\(\text { Now take the roots are }-\alpha,-\beta\)
\(\Rightarrow \text { Sum : }-\alpha-\beta \quad-(\alpha+\beta)=-\frac{5}{3}\)
\(\text { Product : }(-\alpha)(-\beta)=\alpha \beta=\frac{-7}{3}\)
\(\text { The reouired equation }\)
\(x^{2}+\frac{5}{3} x-\frac{7}{3}=0 \)
\(3 x^{2}+5 x-7=0 \)

\(x^{2}+|x-1| =1\)
\(|x-1| =1-x^{2} \)
\(1-x^{2} =x-1 \text { (or) } 1-x^{2} =-x+1 \)
\(x^{2}+x-2 =0 \ x^{2}-x =0 \)
\((x+2)(x-1) =0 \ x(x-1) =0 \)
\(x =-2 \text { (or) } 1 \ x=0 ; x =1 \)
The roots are -2, 0, 1. Since -2 does not satisfy. The roots are 0 and 1.
.'. Number of solution is 2.

\(x^{2}-k x+16=0, a^{} \& b \text { are roots }\)
\(a+b =k ; a b=16 \)
\(a^{2}+b^{2} =(a+b)^{2}-2 a b \)
\(32 =k^{2}-32 \)
\(k^{2} =64 \)
\(k =\pm 8 \Rightarrow k=-8,8 \)

\(2 x^{2}+(a-3) x+3 a-5=0\)
\(\text {Sum }=\frac{-(a-3)}{2} ; \quad \text { Product }=\frac{3 a-5}{2}\)
\(\text {Given they are equal, } \frac{-(a-3)}{2}=\frac{3 a-5}{2}\)
\(-4 a =-8 \)
\(a =2 \)

\(\log _{a} b \log _{b} c \log _{c} a=\log _{a} c \log _{c} a=\log _{a} a=1\)

\(\log _{\sqrt{x}} 0.25 =4 \)
\((\sqrt{x})^{4} =0.25 \)
\((\sqrt{x})^{4} =\frac{1}{4} \)
\(x^{2} =\frac{1}{4} \)
\(\Rightarrow x=\frac{1}{2}=0.5\)

\(\text { In }(-\infty, 2), \frac{|x-2|}{x-2} \text { is negative }\)
\(\operatorname{In}[2, \infty), \frac{|x-2|}{x-2} \geq 0\)

\(|x+2| \leq 9 \)
\( \Rightarrow-9 \leq x+2 \leq 9 \)
\( \Rightarrow-11 \leq x \leq 7 \)
\(x \text { belongs to }[-11,7]\)