Question

How do you solve `(1/x)+1/(x+3)=1/4` using the quadratic formula?

Answer 1

`color(blue)(x=6.772" and" -1.772)`

Explanation:

Consider the left side

Common denominator is `x(x+3)`

So we have:

`((x+3)+x)/(x(x+3))=1/4`

`(2x+3)/(x^2+3x)=1/4`

Multiply both sides by `(x^2+3x)`

`(2x+3)xx(x^2+3x)/(x^2+3x)=(x^2+3x)/4`

but `(x^2+3x)/(x^2+3x)=1`

`(2x+3)=(x^2+3x)/4`

Multiply both sides by 4

`4(2x+3)=x^2+3x`

`8x+12=x^2+3x`

Subtract `8x` and 12 from both sides

`x^2+3x-8x-12=0`

`color(brown)(x^2-5x-12=0)`
`color(brown)("Now we can use the formula")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Standard form `y=ax^2+bx+c`
where `a=1" ; "b=-5" ; "c=-12`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(5+-sqrt((-5)^2-4(1)(-12)))/(2(1))`

`x=(5+-sqrt(25+48))/2`

`x=(5+-sqrt(73))/2" "` Note that 73 is a prime number

`color(blue)(x=6.772" and" -1.772)`