2 years ago

What is the solution of the set?

1 answer:

3 0

$x^2=16-4x\ \ \ |+4x\\\\x^2+4x=16\ \ \ \ |+2^2\\\\\underbrace{x^2+2\cdot x\cdot2+2^2}_{(a+b)^2=a^2+2ab+b^2}=16+2^2\\\\(x+2)^2=20\iff x+2=\pm\sqrt{20}\\\\x+2=-\sqrt{4\cdot5}\ \vee\ x+2=\sqrt{4\cdot5}\ \ \ |-2\\\\x=-2-2\sqrt5\ \vee\ x=-2+2\sqrt5$

Answer: D.

You might be interested in

use the numbers 8, 6, and 2 to create one operation to write an expression that includes one exponent and has a value of 8. use

riadik2000 [5.3K]

Answer:

Answer:

Step-by-step explanation:

1) One solution is to write, as our exponent:

2) Because this is special case of the Exponents Law, valid for every base ≠ 0.

3) Hence, including an exponent the numbers 6, 2 whose value is eight.

Therefore our expression is true

Step-by-step explanation:

4 0