## anonymous one year ago can some one explain the concept of completing the square will fan and medal

1. 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

2. 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 $x^2+ 6x +(\frac{ 6 }{ 2 })^2=-4+(\frac{6}{2})^2$

3. 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 $x=-3 \pm \sqrt{5}$

4. Nnesha

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

5. Nnesha

any question ?

6. anonymous

No, thank u😀