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Question Number 9976 by konen last updated on 20/Jan/17

a^2 +b^2 +4a+6b+13=0    ⇒ a+b=?

$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{4a}+\mathrm{6b}+\mathrm{13}=\mathrm{0}\:\: \\ $$$$\Rightarrow\:\mathrm{a}+\mathrm{b}=? \\ $$

Answered by prakash jain last updated on 20/Jan/17

a^2 +4a+4+b^2 +6b+9=0  (a+2)^2 +(b+3)^2 =0  Since (a+2)^2 ≥0 and (b+3)^2 ≥0  and given that sum   (a+2)^2 +(b+3)^2 =0  ⇒a+2=0 and b+3=0  ⇒a=−2,b=−3  a+b=−5

$${a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{4}+{b}^{\mathrm{2}} +\mathrm{6}{b}+\mathrm{9}=\mathrm{0} \\ $$$$\left({a}+\mathrm{2}\right)^{\mathrm{2}} +\left({b}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{Since}\:\left({a}+\mathrm{2}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\mathrm{and}\:\left({b}+\mathrm{3}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{given}\:\mathrm{that}\:\mathrm{sum}\: \\ $$$$\left({a}+\mathrm{2}\right)^{\mathrm{2}} +\left({b}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}+\mathrm{2}=\mathrm{0}\:\mathrm{and}\:{b}+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{a}=−\mathrm{2},{b}=−\mathrm{3} \\ $$$${a}+{b}=−\mathrm{5} \\ $$

Commented by konen last updated on 20/Jan/17

thank you ⌣

$$\mathrm{thank}\:\mathrm{you}\:\smile \\ $$

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