$x^2+4x+1=0\\\\\underbrace{x^2+2x\cdot2+2^2}_{(x+2)^2}-2^2+1=0\\\\use\ (a+b)^2=a^2+2ab+b^2\\\\(x+2)^2-4+1=0\\\\(x+2)^2-3=0\\\\(x+2)^2=3\iff x+2=\sqrt3\ or\ x+2=-\sqrt3\\\\\boxed{x=\sqrt3-2\ or\ x=-2-\sqrt3}$

7 0

3 years ago

Read 2 more answers

Someone please help!! i will<br> make you a brainliest...i don’t want to fail:(((

Serggg [28]

Answer:

the fisrt one is y=1/2x

the second one is y=x+11

the third one is y= x^3

fourth on is y= ^3 checkmark x

Step-by-step explanation: your welcome

8 0

2 years ago

What is the solution to this inequality?

Evgen [1.6K]

Answer:

Answer: x<7.1

Step-by-step explanation:

${ \tt{4.7 > x - 2 \frac{2}{5} }} \\ \\ { \tt{4.7 > x - \frac{12}{5} }} \\ \\ { \tt{x < 4.7 + \frac{12}{5} }} \\ \\ { \boxed{ \tt{x < 7.1}}}$

6 0

2 years ago