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Question Number 64153 by meme last updated on 14/Jul/19

solve to z^2    x^2 −y^2 =24

$${solve}\:{to}\:{z}^{\mathrm{2}} \:\:\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{24} \\ $$

Answered by mr W last updated on 15/Jul/19

(x−y)(x+y)=24=a×b=1×24=2×12=3×8=4×6  x=((a+b)/2)  y=((b−a)/2)  a and b should be both even or odd,  ⇒a×b=2×12 or 4×6  ⇒x=((2+12)/2)=7, y=((12−2)/2)=5  ⇒x=((4+6)/2)=5, y=((6−4)/2)=1    ⇒(x,y)=(±7,±5) or (±5,±1)

$$\left({x}−{y}\right)\left({x}+{y}\right)=\mathrm{24}={a}×{b}=\mathrm{1}×\mathrm{24}=\mathrm{2}×\mathrm{12}=\mathrm{3}×\mathrm{8}=\mathrm{4}×\mathrm{6} \\ $$$${x}=\frac{{a}+{b}}{\mathrm{2}} \\ $$$${y}=\frac{{b}−{a}}{\mathrm{2}} \\ $$$${a}\:{and}\:{b}\:{should}\:{be}\:{both}\:{even}\:{or}\:{odd}, \\ $$$$\Rightarrow{a}×{b}=\mathrm{2}×\mathrm{12}\:{or}\:\mathrm{4}×\mathrm{6} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}+\mathrm{12}}{\mathrm{2}}=\mathrm{7},\:{y}=\frac{\mathrm{12}−\mathrm{2}}{\mathrm{2}}=\mathrm{5} \\ $$$$\Rightarrow{x}=\frac{\mathrm{4}+\mathrm{6}}{\mathrm{2}}=\mathrm{5},\:{y}=\frac{\mathrm{6}−\mathrm{4}}{\mathrm{2}}=\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\left({x},{y}\right)=\left(\pm\mathrm{7},\pm\mathrm{5}\right)\:{or}\:\left(\pm\mathrm{5},\pm\mathrm{1}\right) \\ $$

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