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anonymous
 3 years ago
solve by completing the square:
4x^27x=3
anonymous
 3 years ago
solve by completing the square: 4x^27x=3

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do u know the method of completing the square first............?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0firstly make the coefficient of first term ie,x^2 unity

Goten77
 3 years ago
Best ResponseYou've already chosen the best response.0if all us fails.. u can even use the quadractic formula

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0NO, the question says to complete the square. @Goten77 if you're good at maths then you should know to never make a rookie mistake of not obeying to the question.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[4x^27x=3\] Divide both sides by 4! \[x^2\frac{7}{4}x=\frac{3}{4}\] Use this property of completing the square. \[x^2bx=c\] \[(x\frac{b}{2})^2=c+(\frac{b}{2})^2\] SO.... \[(x\frac{ 7 }{ 8 })^2=\frac{ 3 }{ 4 }+\frac{ 49 }{ 64 }\] \[(x\frac{ 7 }{ 8 })^2=\frac{ 97 }{ 64 }\] Square root both sides while BEING CAUTIOUS that there must be a negative value as well as a positive. \[x\frac{ 7 }{ 8 }=\frac{ \pm \sqrt{97} }{ 8 }\] \[x=\frac{ \pm \sqrt{97} }{ 8 }+\frac{ 7 }{ 8 }\] \[x=\frac{ 7\pm \sqrt{97} }{ 8 }\]
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