# Which is the best quadratic equation trick

If a quadratic equation only has terms with the variable (mostly \$\$ x \$\$) and should be \$\$ = 0 \$\$, there is a simple trick for solving it.

#### Examples

• \$\$ x ^ 2 + x = 0 \$\$
• \$\$ 2x ^ 2 + 1.5x = 0 \$\$
• \$\$ 2x -x ^ 2 = 0 \$\$

#### Trick: Exclude x = factorize

Solve the equation \$\$ x ^ 2 + 2x = 0 \$\$.

Factor out x. So you get 2 factors. That's why the step is called Factoring.

Exclude \$\$ x ^ 2 + 2x = 0 \$\$ \$\$ | \$\$

\$\$ x (x + 2) = 0 \$\$

There is now a product on the left. A product becomes 0 when one or the other part is 0.

1st case: \$\$ x = 0 \$\$
So 1st solution: \$\$ x_1 = 0 \$\$

2nd case: \$\$ x + 2 = 0 \$\$
So 2nd solution: \$\$ x_2 = -2 \$\$

Solution set: \$\$ L = {0; -2} \$\$

You solve equations that only contain terms with variables by factoring them out. Then apply: A product is 0 if at least one of the factors is 0.

Zero product rate: The product \$\$ a · b \$\$ of two terms a and b is zero if and only if at least one of the factors a or b is zero.
Shorter: From \$\$ a b = 0 \$\$ follows \$\$ a = 0 \$\$ or \$\$ b = 0 \$\$ or both equal 0.

### First transform, then factor

You first transform equations that have a factor before the \$\$ x ^ 2 \$\$.

#### 1st example

\$\$ - 4 \$\$\$\$ x ^ 2-8x = 0 \$\$ \$\$ | \$\$ \$\$: \$\$\$\$ (- 4) \$\$

\$\$ x ^ 2 + 2x = 0 \$\$

\$\$ x (x + 2) = 0 \$\$

1st case: \$\$ x = 0 \$\$
So 1st solution: \$\$ x_1 = 0 \$\$

2nd case: \$\$ x + 2 = 0 \$\$
So 2nd solution: \$\$ x_2 = -2 \$\$

Solution set: \$\$ L = {0; -2} \$\$

#### 2nd example

\$\$ (1) / (3) x ^ 2 = 2x \$\$ \$\$ | · 3 \$\$

\$\$ x ^ 2 = 6x \$\$ | \$\$ - 6x \$\$

\$\$ x ^ 2-6x = 0 \$\$

\$\$ x (x-6) = 0 \$\$

1st case: \$\$ x = 0 \$\$
So 1st solution: \$\$ x_1 = 0 \$\$

2nd case: \$\$ x-6 = 0 \$\$
So 2nd solution: \$\$ x_2 = 6 \$\$

Solution set: \$\$ L = {0; 6} \$\$

If a quadratic equation only contains quadratic and linear terms, you can solve the equation by rearranging and factoring.

The equation is divided by \$\$ - 4 \$\$ so that the coefficient of the square term becomes \$\$ 1 \$\$.

A second solution is possible if you exclude \$\$ - 4x \$\$.
\$\$ - 4x ^ 2-8x = 0 \$\$
\$\$ - 4x (x + 2) = 0 \$\$
1st solution: \$\$ - 4x = 0 \$\$ with \$\$ x_1 = 0 \$\$
2nd solution: \$\$ x + 2 = 0 \$\$ with \$\$ x_2 = -2 \$\$

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### Equations of the form \$\$ x ^ 2 + px = 0 \$\$, \$\$ p in RR \$\$

This is the general solution:

### Solution steps

 Equation: \$\$ x ^ 2 + px = 0 \$\$ Factorize: \$\$ x (x + p) = 0 \$\$ Zero product rate: \$\$ x = 0 \$\$ or \$\$ x + p = 0 \$\$ Solutions: \$\$ x_1 = 0 \$\$ and \$\$ x_2 = -p \$\$ Solution set: \$\$ L = {0 ;-p} \$\$

#### example

 Equation: \$\$ x ^ 2-2.5 x = 0 \$\$ Factorize: \$\$ x (x-2.5) = 0 \$\$ Zero product rate: \$\$ x = 0 \$\$ or \$\$ x-2.5 = 0 \$\$ Solutions: \$\$ x_1 = 0 \$\$ and \$\$ x_2 = 2.5 \$\$\$\$ p = -2.5 \$\$, i.e. \$\$ - p = + 2.5 \$\$ Solution set: \$\$ L = {0; 2.5} \$\$

A Solution of these quadratic equations is always 0!