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a-b-gt-0-1-a-1-b-2-and-a-2-b-2-12-a-b-




Question Number 213621 by hardmath last updated on 10/Nov/24
a,b>0  (1/a) + (1/b) = 2   and   a^2  + b^2  = 12  a + b = ?
$$\mathrm{a},\mathrm{b}>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:=\:\mathrm{2}\:\:\:\mathrm{and}\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:=\:\mathrm{12} \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:=\:? \\ $$
Answered by Ghisom last updated on 10/Nov/24
(1/a)+(1/b)=2 ⇒ 2ab=a+b  a^2 +2ab+b^2 =12+2ab  (a+b)^2 =12+(a+b)  (a+b)^2 −(a+b)−12=0  a+b=(1/2)±(7/2)  a, b >0 ⇒ a+b=4
$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\mathrm{2}\:\Rightarrow\:\mathrm{2}{ab}={a}+{b} \\ $$$${a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} =\mathrm{12}+\mathrm{2}{ab} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{12}+\left({a}+{b}\right) \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} −\left({a}+{b}\right)−\mathrm{12}=\mathrm{0} \\ $$$${a}+{b}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${a},\:{b}\:>\mathrm{0}\:\Rightarrow\:{a}+{b}=\mathrm{4} \\ $$

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