Answer:
https://www.map.mathshell.org/download.php?fileid=1714
Step-by-step explanation:
Go to that website and it should tell you about everything
Strictly speaking, x^2 + 2x + 4 doesn't have solutions; if you want solutions, you must equate <span>x^2 + 2x + 4 to zero:
</span>x^2 + 2x + 4= 0. "Completing the square" seems to be the easiest way to go here:
rewrite x^2 + 2x + 4 as x^2 + 2x + 1^2 - 1^2 = -4, or
(x+1)^2 = -3
or x+1 =i*(plus or minus sqrt(3))
or x = -1 plus or minus i*sqrt(3)
This problem, like any other quadratic equation, has two roots. Note that the fourth possible answer constitutes one part of the two part solution found above.
Answer:
Step-by-step explanation:
y = ax² + bx + c
~~~~~~~
(4, - 2), (8, 6), (2, 0)
a(4²) + b(4) + c = - 2
a(8²) + b(8) + c = 6
a(2²) + b(2) + c = 0
16a + 4b + c = - 2 .............. <em>(1)</em>
64a + 8b + c = 6 ............... <em>(2)</em>
4a + 2b + c = 0 ................. <em>(3)</em>
a = 0.5 ; b = - 4 ; c = 6
<em>y = 0.5x² - 4x + 6</em>
Part A: (0, 6), (6, 0)
Part B: (0,6)
Part C: (6, 0)
Part D: [ 4, ∞ )
<span>x=<span>−<span><span>1<span> or </span></span>x</span></span></span>=<span>−<span>3 i think lol</span></span>