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Question Number 34567 by jordan last updated on 08/May/18

x determinant ((2),())−2x−15=0

$${x}\begin{vmatrix}{\mathrm{2}}\\{}\end{vmatrix}−\mathrm{2}{x}−\mathrm{15}=\mathrm{0} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 08/May/18

x^2 is aleays positive hence ∣x^2 ∣=x^2   x^2 −2x−15=0  (x−5)(x+3)=0  x=5,−3

$${x}^{\mathrm{2}} {is}\:{aleays}\:{positive}\:{hence}\:\mid{x}^{\mathrm{2}} \mid={x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{15}=\mathrm{0} \\ $$$$\left({x}−\mathrm{5}\right)\left({x}+\mathrm{3}\right)=\mathrm{0} \\ $$$${x}=\mathrm{5},−\mathrm{3} \\ $$

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