Solve 4x^2 + 4x = -1. round to the nearest tenth.?

4x^2 + 4x = -1
4x^2 + 4x + 1 = 0
(-4 +/- (sq rt (4^2-4(4)(1))))/2(4)
(-4 +/- (sq rt (0)))/8
-4/8 = -0.5

4x^2 + 4x = -1

4x^2 + 4x +1 = 0.....take factors

(2x + 1)(2x + 1) = 0

This is true when x = -0.5

the answer is a.

[x + (eleven/8)]^2 = 3/4 + 121/sixty 4 [x + (eleven/8)]^2 = 40 8/sixty 4 + 121/sixty 4 [x + (eleven/8)]^2 = 169/sixty 4 sq. root of the two factors, giving x+ eleven/8 = +-13/8 so x = -eleven/8 + 13/8 = 2/8 = a million/4 or x = -eleven/8 - 13/8 = -24/8 = -3

4x^2 + 4x + 1 = 0

x = [ - b ± √ (b ² - 4 a c ) ] / 2 a

x = [ - 4 ± √ (16 - 16 ) ] / 8

x = - 0.5

OPTION a

(a + b)^2 ≡ a^2 + 2ab + b^2

4x^2 + 4x = -1
4x^2 + 4x + 1 = 0
(2x)^2 + 2(2x)(1) + 1^2 = 0
(2x + 1)^2 = 0

2x + 1 = 0
2x = -1
x = -1/2 (-0.5)

∴ x = -1/2 (-0.5)
(answer a)

4x^2 + 4x = -1
4x^2 + 4x +1=0
4x^2/4 + 4x/4 + 1=0
x^2 + x + 1 = 0
x^2/2 + x + 1/2 = 0
x - x = 0+1/2
so, X = -1/2 or -0.5
A. x= -0.5 is your answer

4x^2+4x+1=0
x=(-4 ± √(4^2-4*4*1))/(2*4)=(-4)/8=-0.5

The function has only one solution since 4x^2+4x+1=(2x+1)^2
(:

(2x + 1)^2 = 0
so x = - 1 / 2

4x^2 + 4x + 1 = 0
x^2 + x + 1/4 = 0

D = 1^2 - 4 * 1 * 1/4 = 0

x = (-1 +/- sqrt(0))/2 = -0.5, so answer a

-0.5