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Question Number 112625 by Aina Samuel Temidayo last updated on 09/Sep/20

If a+b=5, a^2 +b^2 =13, the value of  a−b (where a>b) is

$$\mathrm{If}\:\mathrm{a}+\mathrm{b}=\mathrm{5},\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{13},\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$ $$\mathrm{a}−\mathrm{b}\:\left(\mathrm{where}\:\mathrm{a}>\mathrm{b}\right)\:\mathrm{is} \\ $$

Commented byMJS_new last updated on 09/Sep/20

lol...  b=5−a  a^2 +(5−a)^2 =13  a^2 −5a+6=0  (a−3)(a−2)=0  a=2∨a=3∧b=5−a∧a>b ⇒ a=3∧b=2

$$\mathrm{lol}... \\ $$ $${b}=\mathrm{5}−{a} \\ $$ $${a}^{\mathrm{2}} +\left(\mathrm{5}−{a}\right)^{\mathrm{2}} =\mathrm{13} \\ $$ $${a}^{\mathrm{2}} −\mathrm{5}{a}+\mathrm{6}=\mathrm{0} \\ $$ $$\left({a}−\mathrm{3}\right)\left({a}−\mathrm{2}\right)=\mathrm{0} \\ $$ $${a}=\mathrm{2}\vee{a}=\mathrm{3}\wedge{b}=\mathrm{5}−{a}\wedge{a}>{b}\:\Rightarrow\:{a}=\mathrm{3}\wedge{b}=\mathrm{2} \\ $$

Answered by MJS_new last updated on 09/Sep/20

at first sight: a=3∧b=2

$$\mathrm{at}\:\mathrm{first}\:\mathrm{sight}:\:{a}=\mathrm{3}\wedge{b}=\mathrm{2} \\ $$

Commented byAina Samuel Temidayo last updated on 09/Sep/20

Sure but solve it mathematically.

$$\mathrm{Sure}\:\mathrm{but}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{mathematically}. \\ $$

Answered by ajfour last updated on 09/Sep/20

a−b=s > 0  a+b=5  ⇒  (s+5)^2 +(5−s)^2 =52  ⇒  2s^2 = 2   ⇒  s=a−b = 1

$${a}−{b}={s}\:>\:\mathrm{0} \\ $$ $${a}+{b}=\mathrm{5} \\ $$ $$\Rightarrow\:\:\left({s}+\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{5}−{s}\right)^{\mathrm{2}} =\mathrm{52} \\ $$ $$\Rightarrow\:\:\mathrm{2}{s}^{\mathrm{2}} =\:\mathrm{2}\:\:\:\Rightarrow\:\:{s}={a}−{b}\:=\:\mathrm{1} \\ $$

Answered by john santu last updated on 09/Sep/20

(a+b)^2  = 25 ⇒a^2 +b^2 +2ab=25    2ab = 25−13 = 12  ...(i)  a−b = (√((a−b)^2 ))=(√(a^2 +b^2 −2ab))             =(√(13−12)) = (√1) = 1

$$\left({a}+{b}\right)^{\mathrm{2}} \:=\:\mathrm{25}\:\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}=\mathrm{25} \\ $$ $$\:\:\mathrm{2}{ab}\:=\:\mathrm{25}−\mathrm{13}\:=\:\mathrm{12}\:\:...\left({i}\right) \\ $$ $${a}−{b}\:=\:\sqrt{\left({a}−{b}\right)^{\mathrm{2}} }=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{13}−\mathrm{12}}\:=\:\sqrt{\mathrm{1}}\:=\:\mathrm{1} \\ $$

Answered by 1549442205PVT last updated on 09/Sep/20

(a−b)^2 =2(a^2 +b^2 )−(a+b)^2 =2.13−25  =1⇒a−b=1(since a>b)

$$\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)−\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} =\mathrm{2}.\mathrm{13}−\mathrm{25} \\ $$ $$=\mathrm{1}\Rightarrow\mathrm{a}−\mathrm{b}=\mathrm{1}\left(\mathrm{since}\:\mathrm{a}>\mathrm{b}\right) \\ $$

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