Question-212518

Question Number 212518 by efronzo1 last updated on 16/Oct/24

$$\:\:\:\cancel{\underbrace{\downharpoonleft}\underline{}\:} \\ $$

Answered by golsendro last updated on 16/Oct/24

± = (x−a)^2 +a f^(−1) (x)= (√(x−a)) +a ⇔ (x−a)^2 = (√(x−a)) (√(x−a)) ((√((x−a)^3 ))−1)=0 x= a or x=a+1 so if x=5049=a then the other solution

$$\:\:\:\cancel{\underline{\underbrace{\pm}} }=\:\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} +\mathrm{a}\: \\ $$$$\:\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\:\sqrt{\mathrm{x}−\mathrm{a}}\:+\mathrm{a}\: \\ $$$$\:\:\Leftrightarrow\:\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} \:=\:\sqrt{\mathrm{x}−\mathrm{a}} \\ $$$$\:\:\:\:\:\:\:\sqrt{\mathrm{x}−\mathrm{a}}\:\left(\sqrt{\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{3}} }−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{x}=\:\mathrm{a}\:\mathrm{or}\:\mathrm{x}=\mathrm{a}+\mathrm{1} \\ $$$$\:\:\mathrm{so}\:\mathrm{if}\:\mathrm{x}=\mathrm{5049}=\mathrm{a}\:\mathrm{then}\:\mathrm{the}\:\mathrm{other}\:\mathrm{solution} \\ $$$$\:\:\: \\ $$

Answered by mr W last updated on 16/Oct/24

x^2 −2ax+a(a+1)=x x^2 −(2a+1)x+a(a+1)=0 say the other solution is β. 5049+β=2a+1 5049β=a(a+1) 4×5049β=(5049+β)^2 −1 (β−5049)^2 −1=0 ⇒β=5049±1 i.e. the other solution is 5048 or 5050. with a=5048 or 5049.

$${x}^{\mathrm{2}} −\mathrm{2}{ax}+{a}\left({a}+\mathrm{1}\right)={x} \\ $$$${x}^{\mathrm{2}} −\left(\mathrm{2}{a}+\mathrm{1}\right){x}+{a}\left({a}+\mathrm{1}\right)=\mathrm{0} \\ $$$${say}\:{the}\:{other}\:{solution}\:{is}\:\beta. \\ $$$$\mathrm{5049}+\beta=\mathrm{2}{a}+\mathrm{1} \\ $$$$\mathrm{5049}\beta={a}\left({a}+\mathrm{1}\right) \\ $$$$\mathrm{4}×\mathrm{5049}\beta=\left(\mathrm{5049}+\beta\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$\left(\beta−\mathrm{5049}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\beta=\mathrm{5049}\pm\mathrm{1} \\ $$$${i}.{e}.\:{the}\:{other}\:{solution}\:{is}\:\mathrm{5048}\:{or}\:\mathrm{5050}. \\ $$$${with}\:{a}=\mathrm{5048}\:{or}\:\mathrm{5049}. \\ $$

Commented by mr W last updated on 16/Oct/24

Commented by mr W last updated on 16/Oct/24

Commented by golsendro last updated on 16/Oct/24

f : [a,∞)→[a,∞) sir it follow that β = 5059−1 not solution

$$\mathrm{f}\::\:\left[\mathrm{a},\infty\right)\rightarrow\left[\mathrm{a},\infty\right)\:\mathrm{sir} \\ $$$$\:\mathrm{it}\:\mathrm{follow}\:\mathrm{that}\:\beta\:=\:\mathrm{5059}−\mathrm{1}\:\mathrm{not} \\ $$$$\:\mathrm{solution} \\ $$

Commented by mr W last updated on 16/Oct/24

with a=5048 there are two solutions: x=5048, 5049 with a=5049 there are two solutions: x=5049, 5050 what is wrong?

$${with}\:{a}=\mathrm{5048}\:{there}\:{are}\:{two}\:{solutions}: \\ $$$${x}=\mathrm{5048},\:\mathrm{5049} \\ $$$${with}\:{a}=\mathrm{5049}\:{there}\:{are}\:{two}\:{solutions}: \\ $$$${x}=\mathrm{5049},\:\mathrm{5050} \\ $$$${what}\:{is}\:{wrong}? \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply