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john18sanders

  • 3 years ago

solve by completing the square: 4x^2-7x=3

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  1. nitz
    • 3 years ago
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    do u know the method of completing the square first............?

  2. nitz
    • 3 years ago
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    firstly make the coefficient of first term ie,x^2 unity

  3. Goten77
    • 3 years ago
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    |dw:1359101474085:dw|

  4. Goten77
    • 3 years ago
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    if all us fails.. u can even use the quadractic formula

  5. Azteck
    • 3 years ago
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    NO, 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.

  6. Azteck
    • 3 years ago
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    \[4x^2-7x=3\] Divide both sides by 4! \[x^2-\frac{7}{4}x=\frac{3}{4}\] Use this property of completing the square. \[x^2-bx=c\] \[(x-\frac{b}{2})^2=c+(\frac{b}{2})^2\] 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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