Question

How do you solve `4n ^ { 2} + 4n + 1= 0` using the quadratic formula?

Answer 1

`n=-1/2`

Explanation:

The standard form of a `color(blue)"quadratic equation"` is.

`• ax^2+bx+c=0;a!=0`

To find the roots using the `color(blue)"quadratic formula"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(x=(-b+-sqrt(b^2-4ac))/(2a))color(white)(2/2)|)))`

`"here " a=4,b=4" and " c=1`

`rArrn=(-4+-sqrt(4^2-(4xx4xx1)))/(2xx4)`

`color(white)(rArrn)=(-4+-sqrt0)/8=(-4)/8=-1/2larrcolor(red)" repeated root"`

Normally a quadratic equation has 2 solutions ( roots) but in this case has only 1. This informs us that the quadratic touches the x-axis at only 1 point
graph{4x^2+4x+1 [-10, 10, -5, 5]}