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

1 Answer
Apr 11, 2017

Answer:

#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]}