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anonymous

  • one year ago

can some one explain the concept of completing the square will fan and medal

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  1. Nnesha
    • one year ago
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    alright \[\huge\rm Ax^2+Bx+C=0\] we should keep all the terms to the left side except the constant move the constant to the right side ( c is constant in this equation ) \[\large\rm ax^2+bx=-c\] now we can complete the square at left side if there is a common factor(just the coefficient ) at left side take it out we need leading coefficient equal to one so i would divide both sides by a \[\rm x^2+\frac{bx}{a}=-\frac{c}{a}\] \[\rm x^2+\frac{bx}{2a}=-\frac{c}{a}\] now take `half` (b/2) of x term coefficient (middle term in the original equation bx) and the square it add( b/2)^2 both sides \[x^2+\frac{bx}{2a}+(\frac{b}{2a})^2 = \frac{-c}{a}+(\frac{b}{2a})^2\]\[(x+\frac{b}{2a})^2= -\frac{c}{a}+ (\frac{b}{2a})^2\] and then simplify like simple algebra you will get the quadratic formula

  2. Nnesha
    • one year ago
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    let's do an example \[\huge\rm x^2+6x+4=0\] `first move the constant to the right side` (subtract 4 both sides) \[\huge\rm x^2+6x=-4\] take half of x term coefficient (which is 6 in this question ) and square it add square both sides of the equal sign (b/2)^2 \[x^2+ 6x +(\frac{ 6 }{ 2 })^2=-4+(\frac{6}{2})^2\]

  3. Nnesha
    • one year ago
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    now simplify that 6/2 =3 \[\huge\rm x^2+6x+9=-4+9\] \[(x+3)^2 =5 \] now to cancel out the square i willl square root \[\sqrt{(x+3)^2}=\sqrt{5}-------- >x+3=\pm \sqrt{5}\] now we can move 3 to the right side \[x=-3 \pm \sqrt{5}\]

  4. Nnesha
    • one year ago
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    make sense do you understand? sorry i have to go in few mints

  5. Nnesha
    • one year ago
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    any question ?

  6. anonymous
    • one year ago
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    No, thank u😀

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