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: t= -4
Step-by-step explanation:
-6t = 24
t = -4
2 * x/100 = 8
x/100 = 4
x = 400
400%
<h3>
Answer: n = -11</h3>
=========================================================
Explanation:
Since x-2 is a factor of f(x), this means f(2) = 0.
More generally, if x-k is a factor of p(x), then p(k) = 0. This is a special case of the remainder theorem.
So if we plugged x = 2 into f(x), we'd get
f(x) = x^3+x^2+nx+10
f(2) = 2^3+2^2+n(2)+10
f(2) = 8+4+2n+10
f(2) = 2n+22
Set this equal to 0 and solve for n
2n+22 = 0
2n = -22
n = -22/2
n = -11 is the answer
Therefore, x-2 is a factor of f(x) = x^3+x^2-11x+10
Plug x = 2 into that updated f(x) function to find....
f(x) = x^3+x^2-11x+10
f(2) = 2^3+2^2-11(2)+10
f(2) = 8+4-22+10
f(2) = 0
Which confirms our answer.
Answer:
250 is the answers for the question
Step-by-step explanation:
please give me brainlest
