Question

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

Answer 1

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

Answer 2

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
\[x^2+ 6x +(\frac{ 6 }{ 2 })^2=-4+(\frac{6}{2})^2\]

Answer 3

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 \[x=-3 \pm \sqrt{5}\]

Answer 4

make sense do you understand?
sorry i have to go in few mints

Answer 5

any question ?

Answer 6

No, thank u😀