Question

2x^ - 36 = x using quadratic equation

Answer 1

For this case we must solve the following quadratic equation:

`2x ^ 2-36 = x\\2x ^ 2-x-36 = 0`

Where:

`a = 2\\b = -1\\c = -36`

The solution is given by:

`x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}`

Substituting:

`x = \frac {- (- 1) \pm \sqrt {(- 1) ^ 2-4 (2) (- 36)}} {2 (2)}\\x = \frac {1 \pm \sqrt {1 + 288}} {4}\\x = \frac {1 \pm \sqrt {289}} {4}\\x = \frac {1 \pm17} {4}`

We have two roots:

`x_ {1} = \frac {1 + 17} {4} = \frac {18} {4} = \frac {9} {2}\\x_ {2} = \frac {1-17} {4} = \frac {-16} {4} = - 4`

Answer:

`x_ {1} = \frac {9} {2}\\x_ {2} = - 4`