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x-199996-x-2-8x-a10-b-c-a-b-c-




Question Number 9987 by konen last updated on 20/Jan/17
x=199996  x^2 +8x=a10^b −c  ⇒ a+b+c=?
$$\mathrm{x}=\mathrm{199996} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{8x}=\mathrm{a10}^{\mathrm{b}} −\mathrm{c} \\ $$$$\Rightarrow\:\mathrm{a}+\mathrm{b}+\mathrm{c}=? \\ $$
Answered by ridwan balatif last updated on 20/Jan/17
x^2 +8x=a×10^b −c  (x+4)^2 −16=a×10^b −c  (199996+4)^2 −16=a×10^b −c  (2×10^5 )^2 −16=a×10^b −c  4×10^(10) −16=a×10^b −c  a=4,  b=10, c=16  a+b+c=4+10+16=30
$$\mathrm{x}^{\mathrm{2}} +\mathrm{8x}=\mathrm{a}×\mathrm{10}^{\mathrm{b}} −\mathrm{c} \\ $$$$\left(\mathrm{x}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{16}=\mathrm{a}×\mathrm{10}^{\mathrm{b}} −\mathrm{c} \\ $$$$\left(\mathrm{199996}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{16}=\mathrm{a}×\mathrm{10}^{\mathrm{b}} −\mathrm{c} \\ $$$$\left(\mathrm{2}×\mathrm{10}^{\mathrm{5}} \right)^{\mathrm{2}} −\mathrm{16}=\mathrm{a}×\mathrm{10}^{\mathrm{b}} −\mathrm{c} \\ $$$$\mathrm{4}×\mathrm{10}^{\mathrm{10}} −\mathrm{16}=\mathrm{a}×\mathrm{10}^{\mathrm{b}} −\mathrm{c} \\ $$$$\mathrm{a}=\mathrm{4},\:\:\mathrm{b}=\mathrm{10},\:\mathrm{c}=\mathrm{16} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{4}+\mathrm{10}+\mathrm{16}=\mathrm{30} \\ $$

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