Question

$4x^{2} +4x+1=0$

Answer 1

$D=b^2-4ac\\D=4^2-4(4)(1)\\D=16-16\\D=0$

Meaning there's only 1 real root, complete the square and you get the answer.

$(x+y)^2=x^2+2xy+y^2$

We're going to use this formula, so if x² = 4x² then x = 2x

and if y² = 1 then y = 1

Straight to formula

$(2x+1)^2=0\\x=-\frac{1}{2}$

Because (2x+1)² is basically (2x+1)(2x+1), we get the same value of x.

So the answer is x = -1/2

(When D>0, there are 2 real roots.)

(When D=0, there are only 1 real root.)

(When D<0, there are no real roots, but 2 complex roots.)

Answer 2

Answer:

ddo

Step-by-step explanation: