Question Number 158097 by zainaltanjung last updated on 31/Oct/21
Answered by Rasheed.Sindhi last updated on 31/Oct/21
$${x}^{\mathrm{2}} +{px}+\mathrm{12}:{Roots}\:\alpha\:,\:\mathrm{4} \\ $$$$\mathrm{product}\:{of}\:{roots}:\mathrm{4}\alpha=\mathrm{12}\Rightarrow\alpha=\mathrm{3}\:\left({other}\:{root}\right) \\ $$$${sum}\:{of}\:{roots}:\mathrm{4}+\mathrm{3}=−\mathrm{p}\Rightarrow\mathrm{p}=−\mathrm{7} \\ $$$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} −\mathrm{7}{x}+{q}=\mathrm{0} \\ $$$${Roots}:\beta\:\&\beta \\ $$$${sum}\:{of}\:{roots}:\mathrm{2}\beta=−\mathrm{7}\Rightarrow\beta=−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${product}\:{of}\:{roots}:\beta^{\mathrm{2}} ={q}\Rightarrow{q}=\left(−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{49}}{\mathrm{4}} \\ $$
Answered by Rasheed.Sindhi last updated on 31/Oct/21
$$\mathrm{5}{x}^{\mathrm{2}} +\mathrm{13}{x}+{k}=\mathrm{0} \\ $$$${Roots}:\:\alpha,\frac{\mathrm{1}}{\alpha} \\ $$$${product}\:{of}\:{roots}:\:\alpha×\frac{\mathrm{1}}{\alpha}={k}\Rightarrow{k}=\mathrm{1} \\ $$
Commented by otchereabdullai@gmail.com last updated on 31/Oct/21
$$\mathrm{nice}\:\mathrm{one}! \\ $$
Commented by Rasheed.Sindhi last updated on 31/Oct/21
$$\mathcal{T}{han}\mathcal{X}\:{abdullai}! \\ $$