How do you solve $1/18x^2 -1/9x-1=0$ using the quadratic formula?

Question

2053 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-12T06:53:01

$x=1+-sqrt19$

Quadratic formula gives the solution of equation $ax^2+bx+c=0$ as $x=(-b+-sqrt(b^2-4ac))/(2a)$

In $1/18x^2-1/9x-1=0$ , we have $a=1/18$ , $b=-1/9$ and $c=-1$ .

Hence, $x=(-(-1/9)+-sqrt((-1/9)^2-4(1/18)(-1)))/(2xx1/18)$

= $(1/9+-sqrt(1/81+2/9))/(1/9)$

= $(1/9+-sqrt(19/81))/(1/9)$

= $(1+-sqrt19)$

Alternatively, we could have simplified $1/18x^2-1/9x-1=0$ by multiplying each term by $18$ to get

$x^2-2x-18=0$ and then solution would have been

$x=(-(-2)+-sqrt((-2)^2-4xx1xx(-18)))/2$

= $(2+-sqrt(4+72))/2=1+-sqrt19$

How do you solve 1/18x^2 -1/9x-1=0 1/18x^2 -1/9x-1=0 using the quadratic formula?