Respuesta :

CastleRook
Step 1
(x+b/2a)^2-[(b²-4ac)/4a^2]=0
Step 2
(x+b/2a)^2=(b²-4ac)/4a^2
Step 3
getting the square root of both sides we get
x+b/2a=√(b²-4ac)/4a^2
x+b/2a=+/-√(b²-4ac)/2a
Step 4
subtract b/2a from both sides we get:
x+b/2a-b/2a=-b/2a+/-√(b²-4ac)/2a
x=-b/2a+/-√(b²-4ac)/2a
Step 5
Simplifying the above by putting them under the same denominator we get
x=[-b+/-√(b²-4ac)]/2a

Answer:

To complete the derivation of the quadratic equation:

Given:

[tex](x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a^2} = 0[/tex]

Add both sides [tex]\frac{b^2-4ac}{4a^2}[/tex] we have;

[tex](x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}[/tex]

Taking square root both sides we have;

⇒[tex](x+\frac{b}{2a})= \pm \sqrt{\frac{b^2-4ac}{4a^2}}[/tex]

⇒[tex]x+\frac{b}{2a} =\pm \frac{\sqrt{b^2-4ac}}{2a}[/tex]

Subtract [tex]\frac{b}{2a}[/tex] from both sides we have;

[tex]x =-\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}[/tex]

Therefore, complete derivation for the quadratic equation is:

Step 1.

[tex](x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a^2} = 0[/tex]

Step 2.

[tex](x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}[/tex]

Step 3.

[tex](x+\frac{b}{2a})= \pm \sqrt{\frac{b^2-4ac}{4a^2}}[/tex]

Step 4.

[tex]x+\frac{b}{2a} =\pm \frac{\sqrt{b^2-4ac}}{2a}[/tex]

Step 5.

[tex]x =-\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}[/tex]

or

[tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

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