anonymous
  • anonymous
can some one explain the concept of completing the square will fan and medal
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Nnesha
  • Nnesha
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
Nnesha
  • Nnesha
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 \[x^2+ 6x +(\frac{ 6 }{ 2 })^2=-4+(\frac{6}{2})^2\]
Nnesha
  • Nnesha
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 \[x=-3 \pm \sqrt{5}\]

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Nnesha
  • Nnesha
make sense do you understand? sorry i have to go in few mints
Nnesha
  • Nnesha
any question ?
anonymous
  • anonymous
No, thank u😀

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