Question

How do you solve `7x^2 - 4x - 1 = 0` using the quadratic formula?

Answer 1

`y=(2+sqrt11)/7` or `y=(2-sqrt11)/7`

Explanation:

Using quadratic formula the solution of `ax^2+bx+c=0` is given by

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

As in `7x^2-4x-1=0`, `a=7`, `b=-4` and `c=-1`,

solution of `7x^2-4x-1=0` is given by

`y=(-(-4)+-sqrt((-4)^2-4*7*(-1)))/(2*7)`

or `y=(4+-sqrt(16+28))/14=(4+-sqrt44)/14`

Note that `sqrt44=2sqrt11`, hence

`y=(4+-2sqrt11)/14=(2+-sqrt11)/7` i.e.

`y=(2+sqrt11)/7` or `y=(2-sqrt11)/7`

Answer 2

`x=(4+sqrt11)/7,(4-sqrt11)/7`

Explanation:

`color(blue)(7x^2-4x-1=0`

This is a Quadratic equation (in form `ax^2+bx+c=0`)

Use Quadratic formula

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

Remember that `a,bandc` are the coefficients

So,

`color(purple)(a=7,b=-4,c=-1`

`rarrx=(-(-4)+-sqrt(-4^2-4(7)(-1)))/(2(7))`

`rarrx=(4+-sqrt(16-(-28)))/(14)`

`rarrx=(4+-sqrt(16+28))/(14)`

`rarrx=(4+-sqrt(44))/(14)`

`rarrx=(4+-sqrt(4*11))/(14)`

`rarrx=(4+-2sqrt(11))/(14)`

`rarrx=(cancel4^2+-cancel2^1sqrt(11))/(cancel14^7)`

`rarrx=(4+-sqrt11)/7`

Now we have two solutions

`color(indigo)(x=(4+sqrt11)/7`

`color(violet)(x=(4-sqrt11)/(7)`

`color(blue)(ul bar |x=(4+sqrt11)/7,(4-sqrt11)/7|`