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3 years ago

Find a quadratic equation with roots -1+4i and -1-4i.

Mathematics

2 answers:

a_sh-v [17]3 years ago

8 0

Hello,

(x-(-1+4i))(x-(-1-4i))=((x+1)-4i)((x+1+4i)
=((x+1)²+16
=x²+2x+17

Simora [160]3 years ago

3 0

Answer:

$x^2 + 2x + 17$

Step-by-step explanation:

Since, a quadratic equation having roots $\alpha$ and $\beta$ is,

$x^2 - (\alpha+\beta)x + \alpha.\beta$

Here,

$\alpha = -1 + 4i$

$\beta = -1 - 4i$

Hence, the required quadratic equation,

$x^2 - (-1+4i- 1 - 4i)x + (-1+4i)(-1-4i)$

$x^2 + 2x + ( (-1)^2 - (4i)^2)$ ( ∵ a² - b² = (a+b)(a-b) )

$x^2 + 2x + (1 + 16)$ ( ∵ i² = -1 )

$x^2 + 2x + 17$

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3 years ago

Mr. Arellano decided to make cookies for 7th grade students. He was supposed to mix 4 cups of flour, 1.2 cups of sugar, and 2.5

Ahat [919]

Answer:

0.6 cups of sugar and 1.25 cups of chocolate chips

Step-by-step explanation:

We can begin to solve this problem with a simple proportion:

$\frac{4}{2}$ = $\frac{1.2}{x}$ = $\frac{2.5}{y}$

(x = the new # of sugar y = new # of chocolate chips)

In order to keep everything proportional, we must abide by the scale factor shown with the the flour.

Basically, if we're going to use three times the amount of flour we actually need, then we need to triple all our other ingredients.

We can see that the scale factor is one half, so

x = 1.2 * 0.5

and

y = 2.5 * 0.5

This means that:

x = 0.6

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This means that he has to use 0.6 (or 3/5) cups of sugar and 1.25 (or 9/8) cups of chocolate chips.

8 0

3 years ago

Can you help me solve this please step by step

eimsori [14]

First, the values of 2sqrt(3) or 3sqrt(3) are not even twice the length of the given base, 3.

Next, I would say that 6 is the better answer than 12, assuming that the triangle is”drawn-to-scale”.

Finally, I think the hypotenuse is 6.

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2 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

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$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
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$s= \frac{54.81}{2}$
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Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
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We can conclude that the perimeter of our triangle is 140.13 square units.

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