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Question Number 116824 by bemath last updated on 07/Oct/20

$$\mathrm{Given}\mathrm{a}>\mathrm{b}>0,\mathrm{a}\mathrm{\&}\mathrm{b}\mathrm{real}\mathrm{number}\mathrm{such}\mathrm{that}$$ $${\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2=7\mathrm{and}\mathrm{a}-\mathrm{ab}+\mathrm{b}=-1.$$ $$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{a}}^2-{\mathrm{b}}^2$$

Answered by 1549442205PVT last updated on 08/Oct/20

$$\mathrm{From}\mathrm{the}\mathrm{hypothesis}\mathrm{we}\mathrm{have}$$ $$\{\begin{array}{l}{\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2=7\left(1\right)\\ \mathrm{a}-\mathrm{ab}+\mathrm{b}=-1\left(2\right).\end{array}$$ $$\mathrm{a}-\mathrm{ab}+\mathrm{b}=-1\Rightarrow \mathrm{b}=\frac{\mathrm{a}+1}{\mathrm{a}-1}\left(3\right).\mathrm{Replace}$$ $$\mathrm{into}\left(1\right)\mathrm{we}\mathrm{get}{\mathrm{a}}^2+{\left(\frac{\mathrm{a}+1}{\mathrm{a}-1}\right)}^2-\frac{{\mathrm{a}}^2+\mathrm{a}}{\mathrm{a}-1}=7$$ $$\iff {\mathrm{a}}^2+\frac{{\mathrm{a}}^2+2\mathrm{a}+1}{{\mathrm{a}}^2-2\mathrm{a}+1}-\frac{{\mathrm{a}}^2+\mathrm{a}}{\mathrm{a}-1}=7$$ $$\Rightarrow {\mathrm{a}}^4-{2\mathrm{a}}^3+{\mathrm{a}}^2+{\mathrm{a}}^2+2\mathrm{a}+1-({\mathrm{a}}^3-\mathrm{a})$$ $$={7\mathrm{a}}^2-14\mathrm{a}+7\iff {\mathrm{a}}^4-{3\mathrm{a}}^3-{5\mathrm{a}}^2+17\mathrm{a}-6=0$$ $$\iff (\mathrm{a}-3)(\mathrm{a}-2)({\mathrm{a}}^2+2\mathrm{a}-1)=0$$ $$\iff \mathrm{a}\in \{2,3,-1+\sqrt{2}\}.\mathrm{Replace}\mathrm{into}\left(1\right)$$ $$\mathrm{we}\mathrm{get}\mathrm{b}\in \{3,2,-1-\sqrt{2}\}\mathrm{but}\mathrm{since}$$ $$\mathrm{a}>\mathrm{b}>0,\mathrm{we}\mathrm{get}\mathrm{b}=2$$ $$\mathrm{Thus},{\mathrm{a}}^2-{\mathrm{b}}^2=5$$

Commented bybemath last updated on 07/Oct/20

$$\mathrm{sir}\mathrm{given}\mathrm{condition}\mathrm{a}>\mathrm{b}>0\mathrm{why}$$ $${\mathrm{a}}^2-{\mathrm{b}}^2\mathrm{negative}\mathrm{sir}?$$

Commented by1549442205PVT last updated on 07/Oct/20

$$\mathrm{Thank}\mathrm{Sir}.\mathrm{I}\mathrm{mistaked}\mathrm{and}\mathrm{corrected}$$

Answered by MJS_new last updated on 07/Oct/20

$$a^2+b^2=ab+7$$ $$a+b=ab-1$$ $$$$ $${(a+b)}^2=3ab+7$$ $$3ab+7={(ab-1)}^2$$ $${\left(ab\right)}^2-5ab-6=0$$ $$a,b>0\Rightarrow ab=6$$ $$$$ $$a^2+b^2=13$$ $$a+b=5$$ $$$$ $$a^2+b^2={(a-b)}^2+2ab={(a-b)}^2+12$$ $${(a-b)}^2=a^2+b^2-12=1$$ $$\Rightarrow a-b=1$$ $$\Rightarrow a^2-b^2=(a-b)(a+b)=5$$

Commented bybemath last updated on 07/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{prof}$$

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