Completing the square

If you've got a quadratic equation on the form of

$\\ ax^{2}+c=0 \\$

Then you can solve the equation by using the square root of

$\\ x=\pm \sqrt{\frac{-c}{a}} \\$

Example:

$\\ 3x^{2}-243=0 \\\\3x^{2}=243 \\\\x^{2}=\frac{243}{3} \\\\x^{2}=81 \\x=\pm \sqrt{81} \\x=9\: \: or\: \: x=-9 \\$

This method can only be used if b = 0. If we instead have an equation on the form of

$\\ x^{2}+bx=0 \\$

we can't use the square root initially since we do not have c-value. But we can add a constant d to both sides of the equation to get a new equivalent equation that is a perfect square trinomial. Remember that a perfect square trinomial can be written as

$\\ x^{2}+bx + d=\left ( x+d \right )^{2}=0 \\$

This process is called completing the square and the constant d we're adding is

$\\ d=\left (\frac{b}{2} \right )^{2} \\$

Example:

$\\ x^{2}+12x=0 \\$

We begin by finding the constant d that can be used to complete the square.

$\\ d=\left (\frac{b}{2} \right )^{2}=\left ( \frac{12}{2} \right )^{2}=6^{2}=36 \\\\\\ x^{2}+12x+d=0+d\Rightarrow \\x^{2}+12x+36=0+36\Rightarrow \\\\\begin{pmatrix}x+6 \end{pmatrix}^{2}=36 \\\\\\\sqrt{\begin{pmatrix} x+6 \end{pmatrix}^{2}}=\pm \sqrt{36} \\\\\\\begin{matrix} x+6=6\: \: &or\: \: & x+6=-6\\ x=0 & & x=-12 \end{matrix} \\$

The completing the square method could of course be used to solve quadratic equations on the form of

$\\ ax^{2}+bx+c=0 \\$

In this case you will add a constant d that satisfy the formula

$\\ d=\left ( \frac{b}{2} \right )^{2}-c \\$

Video lesson: Solve the equation by completing the squares

$\\x^{2} - 3x - 10=0 \\$

Next Class: Quadratic equations, The quadratic formula

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