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avanturin [10]
3 years ago
5

Write an equation for the polynomial function that has 2 and -3 + 5i as zero

Mathematics
1 answer:
AysviL [449]3 years ago
8 0

Answer:

p(x) = {x}^{3}  + 4{x}^{2}  + 22x- 68

Step-by-step explanation:

We want to write an equation for the polynomial function with roots;

x = 2

and

x =  - 3 + 5i

Note that complex roots come in pairs and in conjugates.

Therefore

x =  - 3 - 5i

is also a root.

By the factor theorem:

(x-2), (x+3+5i), and (x+3-5i) are factors of the polynomial.

Let p(x) be the polynomial, then in factored form.

p(x) = (x - 2)(x + 3 + 5i)(x + 3 - 5i)

We expand now to get:

p(x) = (x - 2)( {x}^{2}  + 3x  -  5ix + 3x + 9 - 15i + 5ix + 15i + 25)

Simplify now

p(x) = (x - 2)( {x}^{2}  + 6x +34)

We expand further;

p(x) = x( {x}^{2}  + 6x + 34) - 2( {x}^{2}  + 6x + 34)

p(x) = {x}^{3}  + 6 {x}^{2}  + 34x - 2 {x}^{2}   - 12x  - 68

p(x) = {x}^{3}  + 4{x}^{2}  + 22x- 68

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<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7BW%7D%20%3D%202" id="TexFormula1" title="\sqrt[4]{W} = 2" alt="\sqrt[4]{W} = 2
solmaris [256]

Answer:

<h2>W = 16</h2>

Step-by-step explanation:

<h3>\sqrt[4]{W}  = 2 </h3>

To find W raise each of the sides of the equation to the power 4 to make W stand alone

That's

<h3>( { \sqrt[4]{W} })^{4}  =  {2}^{4}</h3>

We have

W = 2⁴

We have the final answer as

<h3>W = 16</h3>

Hope this helps you

6 0
3 years ago
Type an equation that can be used to solve for the missing angles. DO NOT SOLVE.
muminat

Answer:

x°+2x°+(x-24)°=180°

Step-by-step explanation:

The sum of of the interior angles of all triangles add up to 180°

Adding x°, 2x°, and (x-24)° <em>will</em> equal 180

Therefore, an equation that can be used to solve for the missing angles is: x°+2x°+(x-24)°=180°

6 0
3 years ago
Use this picture to solve for X
Zepler [3.9K]

Answer:

x=2

Step-by-step explanation:

i would set up a system of equation using the pythagorean theorem:

1^2+x^2=z^2

4^2+x^2=y^2

Add up these equations to get:

17+2x^2=y^2+z^2

But notice that

y^2+z^2=5^2

So you can substitute

17+2x^2=25

2x^2=8

x^2=4

x=2

:)

8 0
3 years ago
Determine the value of x to the nearest thousandth in the equation 8(2)^x+3=48
solong [7]

You probably mean either

8\cdot2^x + 3 = 48

or

8\cdot2^{x+3} = 48

Write 8 = 2³, so that in the first interpretation,

8\cdot2^x = 2^3 \cdot 2^x = 2^{x + 3}

and in the second,

8\cdot2^{x+3} = 2^3 \cdot 2^{x+3} = 2^{x + 6}

Then in the first interpretation, we have

2^{x + 3} + 3 = 48 \implies 2^{x + 3} = 45 \implies x + 3 = \log_2(45) \implies x = \log_2(45) - 3 \approx \boxed{2.492}

Otherwise, the second interpretation gives

2^{x + 6} = 48 \implies x + 6 = \log_2(48) \implies x = \log_2(48) - 6 \approx -0.415

8 0
3 years ago
Solve the following equation by factoring:9x^2-3x-2=0
olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}

Step-by-step explanation:

Original quadratic equation is 9x^{2}-3x-2=0

Divide both sides by 9:

x^{2} - \frac{x}{3} - \frac{2}{9}=0

Add \frac{2}{9} to both sides to get rid of the constant on the LHS

x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}  ==> x^{2} - \frac{x}{3}=\frac{2}{9}

Add \frac{1}{36}  to both sides

x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}

This simplifies to

x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = \frac{1}{6}\right) we can see that

\left(x - \frac{1}{6}\right)^2 = x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}

So

\left(x - \frac{1}{6}\right)^2=\frac{1}{4}

Taking square roots on both sides

\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}

So the two roots or solutions of the equation are

x - \frac{1}{6}=-\sqrt{\frac{1}{4}}  and x - \frac{1}{6}=\sqrt{\frac{1}{4}}

\sqrt{\frac{1}{4}} = \frac{1}{2}

So the two roots are

x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}

and

x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}

7 0
2 years ago
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