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anonymous
 5 years ago
solve the equation
x+1=sqrt/19x
anonymous
 5 years ago
solve the equation x+1=sqrt/19x

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So I think you are trying to solve the equation: \[x+1=\sqrt{19x}\] Right? If not, let me know... To solve this equation you need to square both sides... the square cancels out the square root but you have to be careful! Squaring both sides of an equation can give solutions that aren't of the original equation. These are called extraneous solutions... \[(x+1)^2=(\sqrt{19x})^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont kno how to write it step by step i dont get it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0O.k., \[(x+1)^2= (\sqrt{19x})^2\] \[(x+1)(x+1)=19x\] \[x^2+x+x+1=19x\] \[x^2+2x+1=19x\] 19 +x 19 +x \[x^2+3x18=0\] (x3)(x+6)=0 so either x = 3 or x = 6 Now you have to check your answer in the original equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if x = 3, \[x+1 = \sqrt{19x}\] becomes \[(3)+1 = \sqrt{19(3)}\] which reduces to \[4 = \sqrt{16}\] which is true so x = 3 is a solution. if x = 6, \[x+1 = \sqrt{19x}\] becomes \[(6)+1 = \sqrt{19(6)}\] which reduces to \[x+1 = \sqrt{19x}\] which becomes \[5 = \sqrt{25}\] which is 5 = 5 which isn't true so x is not equal to 6 Therefore your only solution is x = 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that enough steps? :)
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