QUADRATIC EQUATION BY QUADRATIC FORMULA

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The quadratic formula is:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Wondering how we got these formulas? 
Look at this!!!

The general form of a quadratic equation is:

$$Ax^2+Bx+C=0$$

Divide through by A
$\frac{Ax^2}{A}+\frac{Bx}{A}+\frac{C}{A}=\frac{0}{A}$
$x^2+\frac{Bx}{A}+\frac{C}{A}=0$

By completing the square
$x^2+\frac{Bx}{A}+\frac{C}{A}=0$
$x^2+\frac{Bx}{A}=-\frac{C}{A}$

Add the square half of the coefficient of x to both sides.
$x^2+\frac{Bx}{A}+(\frac{B}{2A})^2=-\frac{C}{A}+(\frac{B}{2})^2$
$x^2+\frac{Bx}{A}+\frac{B^2}{4A^2}=-\frac{C}{A}+\frac{B^2}{4A^2}$

Solving the right-hand side
$x^2+\frac{Bx}{A}+\frac{B^2}{4A^2}=\frac{-4AC+B^2}{4A^2}$

$(x^2+\frac{Bx}{A}+\frac{B^2}{4A^2})$ is  $(x+\frac{b}{2A})^2$.

$(x+\frac{B}{2A})^2=\frac{-4AC+B^2}{4A^2}$

Taking the square root of both sides

$\sqrt{(x+\frac{B}{2A})^2}=\sqrt{\frac{-4AC+B^2}{4A^2}}$
$x+\frac{B}{2A}=\pm\sqrt{\frac{-4AC+B^2}{4A^2}}$

$x+\frac{B}{2A}=\pm\sqrt{\frac{B^2-4AC}{4A^2}}$

For the sake of simplicity, 
$x+\frac{B}{2A}=\pm\frac{\sqrt{B^2-4AC}}{\sqrt{(2A)^2}}$
$x+\frac{B}{2A}=\pm\frac{\sqrt{B^2-4AC}}{\sqrt{(2A)^2}}$
$x+\frac{B}{2A}=\pm\frac{\sqrt{B^2-4AC}}{2A}$

Let's re-arrange the equation so that $x$ will be on one side while another variable will be placed on the other side.
$x=-\frac{B}{2A}\pm\frac{\sqrt{B^2-4AC}}{2A}$
$x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$

If you're still confused about how we came up with the quadratic formula, check out my post on how to solve a quadratic equation by completing the square

Having proved the quadratic equation, let's look at more realistic illustrations

Example 1
Solve $4x^2+15x+9=0$ using the quadratic formula.

Solution
Here $A=4$, $B=15$, $C=9$
Remember that $x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$
$x=\frac{-(15)\pm\sqrt{(15)^2-4(4)(9)}}{2(4)}$
$x=\frac{-15)\pm\sqrt{225-144}}{8}$
$x=\frac{-15\pm\sqrt{81}}{8}$
$x=\frac{-15\pm9}{8}$
$x=\frac{-15-9}{8}$ or $\frac{-15+9}{8}$
$x=\frac{-24}{8}$ or $\frac{-6}{8}$
$x=-3$ or $-0.75$

Example 2
Using the quadratic formula, solve $12x^2+11x+2=0$

Solution
Here, $A=12$, $B=11$, $C=2$
$x=\frac{-(11)\pm\sqrt{(11)^2-4(12)(2)}}{2(12)}$
$x=\frac{-11\pm\sqrt{121-96}}{24}$
$x=\frac{-11\pm\sqrt{25}}{24}$
$x=\frac{-11\pm5}{24}$
$x=\frac{-11+5}{24}$ or $\frac{-11-5}{24}$
$x=\frac{-1}{4}$ or $\frac{-2}{3}$
$x=-0.25$ or $-0.67$

Example 3
Compute $9x^2+3x-2=0$ using the quadratic formula

Solution
$A=9$, $B=3$, $C=-2$
$x=\frac{-(3)\pm\sqrt{(3)^2-4(9)(-2)}}{2(9)}$
$x=\frac{-3\pm\sqrt{9+72}}{18}$
$x=\frac{-3\pm\sqrt{81}}{18}$
$x=\frac{-3\pm9}{18}$
$x=\frac{-3+9}{18}$ or $\frac{-3-9}{18}$
$x=\frac{1}{3}$ or $\frac{-2}{3}$
$x=0.333$ or $-0.667$

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Example 4
Solve $3x^2-5x-2=0$ using the quadratic formula

Solution
Here $A=3$, $B=-5$, $C=-2$
$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(3)(-2)}}{2(3)}$
$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(3)(-2)}}{2(3)}$
$x=\frac{+5\pm\sqrt{25+24}}{6}$
$x=\frac{+5\pm\sqrt{49}}{6}$
$x=\frac{+5\pm7}{6}$
Splitting the $\pm$
$x=\frac{5+7}{6}$ or $\frac{5-7}{6}$
$x=2$ or $-0.333$

The quadratic formula has been used to solve quadratic equations. 

If you're having trouble understanding this, I recommend going over it again.


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