Question

How do you solve `2x ^ { 2} + 23x + 63= 0`?

Answer 1

Find factors and let each one be equal to `0`

`x = -9/2 or x = -7`

Explanation:

When solving a quadratic equation, try to factorise the trinomial first:

`2x^2 +23x +63 =0`

In this case, find factors of `2 and 63` such that their products ADD to `23`

`color(white)(xxxxx)2 and 63`
`color(white)(x.xxx)darr and darr`

`color(white)(xxxxx)2color(white)(wwwww)9 " "rarr 1 xx 9 = 9`
`color(white)(xxxxx)1color(white)(wwwww)7 " "rarr 2xx 7 = ul14`
`color(white)(xxxxxx.xxxxxxxxxxxxxxxx)23`

`2x^2 +23x +63 =0`
`(2x+9)(1x+7)=0`

Either bracket can be equal to `0`

If `2x+9 = 0 " "rarr x = -9/2`

If `x+7 =0" "rarr x = -7`