Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 32357 by .none. last updated on 23/Mar/18

x^2 +ax−24=0  root is integer  a   range    i cant speak english well. sorry

$${x}^{\mathrm{2}} +{ax}−\mathrm{24}=\mathrm{0} \\ $$$${root}\:{is}\:{integer} \\ $$$${a}\:\:\:{range} \\ $$$$ \\ $$$${i}\:{cant}\:{speak}\:{english}\:{well}.\:{sorry} \\ $$

Answered by mrW2 last updated on 23/Mar/18

let p, q be the integer roots.  pq=−24  a=−(p+q)    if pq=−24=−1×24:  p=−1, q=24⇒a=−(−1+24)=−23  or  p=1, q=−24⇒a=−(1−24)=23  or  p=24, q=−1⇒a=−(24−1)=−23  or  p=−24, q=1⇒a=−(−24+1)=23  ⇒a=±23  similarly,  24=2×12⇒a=±10  24=3×8⇒a=±5  24=4×6⇒a=±2

$${let}\:{p},\:{q}\:{be}\:{the}\:{integer}\:{roots}. \\ $$$${pq}=−\mathrm{24} \\ $$$${a}=−\left({p}+{q}\right) \\ $$$$ \\ $$$${if}\:{pq}=−\mathrm{24}=−\mathrm{1}×\mathrm{24}: \\ $$$${p}=−\mathrm{1},\:{q}=\mathrm{24}\Rightarrow{a}=−\left(−\mathrm{1}+\mathrm{24}\right)=−\mathrm{23} \\ $$$${or} \\ $$$${p}=\mathrm{1},\:{q}=−\mathrm{24}\Rightarrow{a}=−\left(\mathrm{1}−\mathrm{24}\right)=\mathrm{23} \\ $$$${or} \\ $$$${p}=\mathrm{24},\:{q}=−\mathrm{1}\Rightarrow{a}=−\left(\mathrm{24}−\mathrm{1}\right)=−\mathrm{23} \\ $$$${or} \\ $$$${p}=−\mathrm{24},\:{q}=\mathrm{1}\Rightarrow{a}=−\left(−\mathrm{24}+\mathrm{1}\right)=\mathrm{23} \\ $$$$\Rightarrow{a}=\pm\mathrm{23} \\ $$$${similarly}, \\ $$$$\mathrm{24}=\mathrm{2}×\mathrm{12}\Rightarrow{a}=\pm\mathrm{10} \\ $$$$\mathrm{24}=\mathrm{3}×\mathrm{8}\Rightarrow{a}=\pm\mathrm{5} \\ $$$$\mathrm{24}=\mathrm{4}×\mathrm{6}\Rightarrow{a}=\pm\mathrm{2} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com