Question Number 202198 by MATHEMATICSAM last updated on 22/Dec/23 $$\mathrm{If}\:\alpha,\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:−\:{b}\:=\:\mathrm{0}\: \\ $$$$\mathrm{and}\:\gamma,\:\delta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{c}\:=\:\mathrm{0}\: \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:\frac{\alpha\:−\:\gamma}{\beta\:−\:\gamma}\:=\:\frac{\beta\:−\:\delta}{\alpha\:−\:\delta}\:\:. \\ $$ Answered by MM42 last updated on 22/Dec/23…
Question Number 202193 by MATHEMATICSAM last updated on 22/Dec/23 $$\mathrm{If}\:\alpha,\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:=\:\:\mathrm{0}\: \\ $$$$\mathrm{and}\:\alpha\:+\:\delta,\:\beta\:+\:\delta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:+\:{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\: \\ $$$${a}^{\mathrm{2}} \:−\:{p}^{\mathrm{2}} \:=\:\mathrm{4}\left({b}\:−\:{q}\right). \\ $$ Answered by Rasheed.Sindhi…
Question Number 202184 by mr W last updated on 22/Dec/23 $${with}\:{f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{30}\:{and}\:{x}\in{R} \\ $$$${solve}\:{f}\left({f}\left({f}\left({f}\left({f}\left({x}\right)\right)\right)\right)\right)=\mathrm{0} \\ $$ Answered by cortano12 last updated on 22/Dec/23 $$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{6}\right)^{\mathrm{2}} −\mathrm{6}\:\Rightarrow\mathrm{x}=\sqrt{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{6}}−\mathrm{6}…
Question Number 202208 by hardmath last updated on 22/Dec/23 Commented by mr W last updated on 22/Dec/23 $${which}\:{country}\:{are}\:{you}\:{from}? \\ $$$${internationally}\:{the}\:{ABO}\:{blood} \\ $$$${group}\:{system}\:{is}\:{common},\:{see}\:{below}. \\ $$$${but}\:{you}\:{are}\:{talking}\:{about}\:{blood}\: \\…
Question Number 202129 by MATHEMATICSAM last updated on 21/Dec/23 $$\mathrm{If}\:{p}\:=\:\mathrm{9999}\:\mathrm{then}\:\frac{\mathrm{4}{p}^{\mathrm{3}} \:−\:{p}}{\left(\mathrm{2}{p}\:+\:\mathrm{1}\right)\left(\mathrm{6}{p}\:−\:\mathrm{3}\right)}\:=\:? \\ $$ Answered by AST last updated on 21/Dec/23 $$\frac{{p}\left(\mathrm{2}{p}−\mathrm{1}\right)\left(\mathrm{2}{p}+\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{2}{p}+\mathrm{1}\right)\left(\mathrm{2}{p}−\mathrm{1}\right)}=\frac{{p}=\mathrm{9999}}{\mathrm{3}}=\mathrm{3333} \\ $$ Terms of…
Question Number 202120 by MATHEMATICSAM last updated on 21/Dec/23 $$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}.\:\mathrm{If}\:\:\frac{\mathrm{1}\:−\:\alpha}{\alpha}\:\mathrm{and}\:\frac{\mathrm{1}\:−\:\beta}{\beta}\:\mathrm{are} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{a}_{\mathrm{1}} {x}^{\mathrm{2}} \:+\:{b}_{\mathrm{1}} {x}\:+\:{c}_{\mathrm{1}} \:=\:\mathrm{0}.\:\mathrm{If} \\ $$$$\frac{{b}_{\mathrm{1}} }{{a}_{\mathrm{1}} }\:=\:{k}\:+\:\frac{{b}}{{c}}\:\mathrm{then}\:{k}\:=\:? \\ $$…
Question Number 202116 by Calculusboy last updated on 21/Dec/23 $$\sqrt{\mathrm{3}^{\boldsymbol{{x}}} }\:+\mathrm{1}=\mathrm{2}^{\boldsymbol{{x}}} \:\:\:\boldsymbol{{find}}\:\boldsymbol{{x}} \\ $$ Answered by Frix last updated on 21/Dec/23 $$\mathrm{Obviously}\:{x}=\mathrm{2} \\ $$$$\sqrt{\mathrm{3}^{\mathrm{2}} }+\mathrm{1}=\sqrt{\mathrm{9}}+\mathrm{1}=\mathrm{3}+\mathrm{1}=\mathrm{4}=\mathrm{2}^{\mathrm{2}}…
Question Number 202146 by hardmath last updated on 21/Dec/23 $$ \\ $$There are three children in a family. Two of the children have blood group…
Question Number 202114 by MATHEMATICSAM last updated on 21/Dec/23 $$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{and}\:\alpha\:+\:{k}\:\mathrm{and}\:\beta\:+\:{k}\:\:\mathrm{are}\: \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{lx}^{\mathrm{2}} \:+\:{mx}\:+\:{n}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:{k}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{b}}{{a}}\:−\:\frac{{m}}{{l}}\right). \\ $$ Answered by aleks041103 last updated…
Question Number 202109 by hardmath last updated on 20/Dec/23