X^2+4+10=0
x=(-4(+/-)root16-40)/2, so we now know that the zeros are imaginary, because you can't square root a negative number and 16-40 is -24
so the two roots are…
-2+iroot6 and -2-iroot6
Answer: a) zeros: x = {0, 4, -2}
b) as x → ∞, y → ∞
as x → -∞, y → ∞
<u>Step-by-step explanation:</u>
I think you mean (a) find the zeros and (b) describe the end behavior
(a) Find the zeros by setting each factor equal to zero and solving for x:
x (x - 4) (x + 2)⁴ = 0
- x = 0 Multiplicity of 1 --> odd multiplicity so it crosses the x-axis
- x = 4 Multiplicity of 1 --> odd multiplicity so it crosses the x-axis
- x = -2 <u>Multiplicity of 4 </u> --> even multiplicity so it touches the x-axis
Degree = 6
(b) End behavior is determined by the following two criteria:
- Sign of Leading Coefficient (Right side): Positive is ↑, Negative is ↓
- Degree (Left side): Even is same direction as right side, Odd is opposite direction of right side
Sign of the leading coefficient is Positive so right side goes UP
as x → ∞, y → ∞
Degree of 6 is Even so Left side is the same direction as right (UP)
as x → -∞, y → ∞
Answer:
Step-by-step explanation:
$ 449.75
Multiply first then round
Let's represent the two numbers by x and y. Then xy=60. The smaller number here is x=y-7.
Then (y-7)y=60, or y^2 - 7y - 60 = 0. Use the quadratic formula to (1) determine whether y has real values and (2) to determine those values if they are real:
discriminant = b^2 - 4ac; here the discriminant is (-7)^2 - 4(1)(-60) = 191. Because the discriminant is positive, this equation has two real, unequal roots, which are
-(-7) + sqrt(191)
y = -------------------------
-2(1)
and
-(-7) - sqrt(191)
y = ------------------------- = 3.41 (approximately)
-2(1)
Unfortunately, this doesn't make sense, since the LCM of two numbers is generally an integer.
Try thinking this way: If the LCM is 60, then xy = 60. What would happen if x=5 and y=12? Is xy = 60? Yes. Is 5 seven less than 12? Yes.
