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x-2-ax-24-0-root-is-integer-a-range-i-cant-speak-english-well-sorry-




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

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