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Question Number 93439 by i jagooll last updated on 13/May/20

Answered by niroj last updated on 13/May/20

$${\mathrm{a}}^2+{\mathrm{b}}^2=2\mathrm{a}-\frac{8\mathrm{b}+1}{{4\mathrm{b}}^2}$$$${6\mathrm{b}}^2+5\mathrm{a}=?$$$$\mathrm{Solution}:$$$${\mathrm{a}}^2+{\mathrm{b}}^2=2\mathrm{a}-\frac{8\mathrm{b}+1}{{4\mathrm{b}}^2}$$$${\mathrm{b}}^2-2\mathrm{a}=-{\mathrm{a}}^2-\frac{8\mathrm{b}+1}{{4\mathrm{b}}^2}$$$$\mathrm{Added}\mathrm{both}\mathrm{sith}\mathrm{by}7\mathrm{a}+{5\mathrm{b}}^2$$$$7\mathrm{a}+{5\mathrm{b}}^2+{\mathrm{b}}^2-2\mathrm{a}=7\mathrm{a}+{5\mathrm{b}}^2-\frac{8\mathrm{b}+1}{{4\mathrm{b}}^2}-{\mathrm{a}}^2$$$${6\mathrm{b}}^2+5\mathrm{a}=-{\mathrm{a}}^2+{5\mathrm{b}}^2+7\mathrm{a}-\frac{2}{\mathrm{b}}-\frac{1}{{4\mathrm{b}}^2}//.$$

Commented by i jagooll last updated on 13/May/20

but the answer is in number sir

Commented by niroj last updated on 13/May/20

$$Accordingyour{que}^{n}itcomes$$$$ifwantinnumericalanswer$$$$Plsmakesure{que}^{n}isclearmydear.$$

Commented by i jagooll last updated on 13/May/20

yes sir. written in the book so the problem

Answered by 1549442205 last updated on 13/May/20

$$\mathrm{I}\mathrm{think}\mathrm{that}\mathrm{this}\mathrm{problem}\mathrm{has}\mathrm{not}\mathrm{numberal}$$$$\mathrm{result}.\mathrm{Inddeed},\mathrm{it}\mathrm{is}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{that}$$$$\mathrm{a}=\frac{1\pm \sqrt{-{({2\mathrm{b}}^2-1)}^2-8\mathrm{b}}}{2\mathrm{b}}(\mathrm{we}\mathrm{consider}\mathrm{as}\mathrm{a}$$$$\mathrm{quadratic}\mathrm{equation}\mathrm{of}\mathrm{a}).\mathrm{Hence}{6\mathrm{b}}^2+5\mathrm{a}\mathrm{can}$$$$\mathrm{receive}\mathrm{a}\mathrm{lot}\mathrm{of}\mathrm{different}\mathrm{values}.\mathrm{We}\mathrm{only}\mathrm{need}$$$$\mathrm{a}\mathrm{condition}\mathrm{that}{({2\mathrm{b}}^2-1)}^2+8\mathrm{b})⩽0(\mathrm{for}△⩾0)$$$$$$

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