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a-2-b-73-b-2-a-73-find-a-b-




Question Number 199112 by hardmath last updated on 28/Oct/23
 { ((a^2  − b = 73)),((b^2  − a = 73)) :}     find: a,b = ?
$$\begin{cases}{\mathrm{a}^{\mathrm{2}} \:−\:\mathrm{b}\:=\:\mathrm{73}}\\{\mathrm{b}^{\mathrm{2}} \:−\:\mathrm{a}\:=\:\mathrm{73}}\end{cases}\:\:\:\:\:\mathrm{find}:\:\mathrm{a},\mathrm{b}\:=\:? \\ $$
Answered by mr W last updated on 28/Oct/23
(i)−(ii):  a^2 −b^2 +a−b=0  (a+b+1)(a−b)=0  case 1: a−b=0   a=b  b^2 −b−73=0  ⇒b=((1±(√(293)))/2)=a  case 2: a+b+1=0  a=−b−1  b^2 +b−72=0  b=((−1±17)/2)=8 or −9  a=−9 or 8
$$\left({i}\right)−\left({ii}\right): \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{a}−{b}=\mathrm{0} \\ $$$$\left({a}+{b}+\mathrm{1}\right)\left({a}−{b}\right)=\mathrm{0} \\ $$$${case}\:\mathrm{1}:\:{a}−{b}=\mathrm{0}\: \\ $$$${a}={b} \\ $$$${b}^{\mathrm{2}} −{b}−\mathrm{73}=\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}\pm\sqrt{\mathrm{293}}}{\mathrm{2}}={a} \\ $$$${case}\:\mathrm{2}:\:{a}+{b}+\mathrm{1}=\mathrm{0} \\ $$$${a}=−{b}−\mathrm{1} \\ $$$${b}^{\mathrm{2}} +{b}−\mathrm{72}=\mathrm{0} \\ $$$${b}=\frac{−\mathrm{1}\pm\mathrm{17}}{\mathrm{2}}=\mathrm{8}\:{or}\:−\mathrm{9} \\ $$$${a}=−\mathrm{9}\:{or}\:\mathrm{8} \\ $$
Commented by hardmath last updated on 28/Oct/23
thankyou professor
$$\mathrm{thankyou}\:\mathrm{professor} \\ $$
Answered by Frix last updated on 28/Oct/23
x^2 −y=z  y^2 −x=z  ⇒  y=z−x^2   x^4 −2zx^2 −x+z^2 −z=0  (x^2 −x−z)(x^2 +x−z+1)=0  ⇒  x=y=((1±(√(4z+1)))/2)  x=((−1±(√(4z−3)))/2)∧y=((−1∓(√(4z−3)))/2)
$${x}^{\mathrm{2}} −{y}={z} \\ $$$${y}^{\mathrm{2}} −{x}={z} \\ $$$$\Rightarrow \\ $$$${y}={z}−{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}{zx}^{\mathrm{2}} −{x}+{z}^{\mathrm{2}} −{z}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −{x}−{z}\right)\left({x}^{\mathrm{2}} +{x}−{z}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}={y}=\frac{\mathrm{1}\pm\sqrt{\mathrm{4}{z}+\mathrm{1}}}{\mathrm{2}} \\ $$$${x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{4}{z}−\mathrm{3}}}{\mathrm{2}}\wedge{y}=\frac{−\mathrm{1}\mp\sqrt{\mathrm{4}{z}−\mathrm{3}}}{\mathrm{2}} \\ $$

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