If-a-b-5-a-2-b-2-13-the-value-of-a-b-where-a-gt-b-is-

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 by MJS_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

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

Commented by Aina Samuel Temidayo last updated on 09/Sep/20

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

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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