Question-211393

Question Number 211393 by BaliramKumar last updated on 08/Sep/24

Commented by BaliramKumar last updated on 08/Sep/24

a^2 −b^2 = 25 a+b = 5 (a−b)(a+b) = 25 (a−b)5 = 25 a−b = 5.........(i) a+b = 5 ..........(ii) a = 5 b = 0 a^2 + ab + b^2 = 5^2 + 5×0 + 0^2 = 25 Is it possible?

$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} \:=\:\mathrm{25}\:\:\:\:\:\:\:\:\:\:\:{a}+{b}\:=\:\mathrm{5} \\ $$$$\left({a}−{b}\right)\left({a}+{b}\right)\:=\:\mathrm{25} \\ $$$$\left({a}−{b}\right)\mathrm{5}\:=\:\mathrm{25} \\ $$$${a}−{b}\:=\:\mathrm{5}………\left({i}\right) \\ $$$${a}+{b}\:=\:\mathrm{5}\:……….\left({ii}\right) \\ $$$${a}\:=\:\mathrm{5}\:\:\:\:\:\:\:\:{b}\:=\:\mathrm{0} \\ $$$${a}^{\mathrm{2}} +\:{ab}\:+\:{b}^{\mathrm{2}} \:=\:\mathrm{5}^{\mathrm{2}} \:+\:\mathrm{5}×\mathrm{0}\:+\:\mathrm{0}^{\mathrm{2}} \:=\:\mathrm{25} \\ $$$${Is}\:\:{it}\:{possible}? \\ $$

Answered by mehdee1342 last updated on 08/Sep/24

(a−b)(a+b)=25⇒a−b=5 (a−b)(a^2 +ab+b^2 )=625 ⇒a^2 +ab+b^2 =125 ✓

$$\left({a}−{b}\right)\left({a}+{b}\right)=\mathrm{25}\Rightarrow{a}−{b}=\mathrm{5} \\ $$$$\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)=\mathrm{625} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} =\mathrm{125}\:\checkmark \\ $$$$ \\ $$

Answered by som(math1967) last updated on 08/Sep/24

a−b=((25)/5)=5 a^3 −b^3 =625 a^2 +ab+b^2 =((625)/5)=125 (b)✓

$$\:{a}−{b}=\frac{\mathrm{25}}{\mathrm{5}}=\mathrm{5} \\ $$$$\:{a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\mathrm{625} \\ $$$$\:{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} =\frac{\mathrm{625}}{\mathrm{5}}=\mathrm{125} \\ $$$$\left({b}\right)\checkmark \\ $$

Commented by som(math1967) last updated on 08/Sep/24

a=5,b=0 then a^3 −b^3 =125 but given a^3 −b^3 =625 ??

$$\:{a}=\mathrm{5},{b}=\mathrm{0}\:{then}\:{a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\mathrm{125} \\ $$$${but}\:{given}\:{a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\mathrm{625}\:\:?? \\ $$

Answered by Frix last updated on 08/Sep/24

We know from Linear Algebra: 3 equations 2 unknowns ⇒ Solve 2 equations and test the 3^(rd) Or graph it: (1) f: b=5−a (2) g: b=±(√(a^2 −25)) (3) h: b=((a^3 −625))^(1/3) ⇒ no common intersection.

$$\mathrm{We}\:\mathrm{know}\:\mathrm{from}\:\mathrm{Linear}\:\mathrm{Algebra}: \\ $$$$\:\:\:\:\:\mathrm{3}\:\mathrm{equations} \\ $$$$\:\:\:\:\:\mathrm{2}\:\mathrm{unknowns} \\ $$$$\:\:\:\:\:\Rightarrow \\ $$$$\:\:\:\:\:\mathrm{Solve}\:\mathrm{2}\:\mathrm{equations}\:\mathrm{and}\:\mathrm{test}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \\ $$$$ \\ $$$$\mathrm{Or}\:\mathrm{graph}\:\mathrm{it}: \\ $$$$\left(\mathrm{1}\right)\:{f}:\:{b}=\mathrm{5}−{a} \\ $$$$\left(\mathrm{2}\right)\:{g}:\:{b}=\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{25}} \\ $$$$\left(\mathrm{3}\right)\:{h}:\:{b}=\sqrt[{\mathrm{3}}]{{a}^{\mathrm{3}} −\mathrm{625}} \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{common}\:\mathrm{intersection}. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply