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3 years ago

3x²+8x+4 how to solve this

2 answers:

7 0

3x² + 8x + 4
First, i divide the equation into two parenthesis, so that the first parts of both multiply to make the first term, 3x². 3x * x = 3x²
(3x + )(x + )
Then I find two numbers that multiply to make 4, which are 1 and 4, or 2 and 2.
Our options are:
(3x + 1)(x + 4)
(3x + 4)(x + 1)
(3x + 2)(3x +2)

To figure out which one to use, I'm just going to FOIL them all.
(3x + 1)(x + 4) = 3x² + 1x + 12x + 4 = 3x² + 13x + 4
(3x + 4)(x + 1) = 3x² + 4x + 3x + 4 = 3x² + 7x + 4
(3x + 2)(x +2) = 3x² + 2x + 6x + 4 = 3x² + 8x + 4

The factored form is:
(3x + 2)(x + 2)

To solve for the roots ( where the graph crosses the x axis, where y = 0) we set the equation equal to 0:

(3x + 2)(x + 2) = 0
The zero product property says that anything times 0 is 0, so we set each individual part equal to 0 and solve for the two roots.

3x + 2 = 0
3x = -2
x = -2/3

x + 2 = 0
x = -2

lisov135 [29]3 years ago

3 0

In this type of equation you could use different methods. But I choose to do Quadratic Method.

a= 3, b= 8, and c= 4.

X= -(8)+_SQUARE ROOT (8)^2-4(3)(4)/2(3)

X= -8+_SQUARE ROOT16/6 Square Root of 16 is 4

X= -8+_4/6

X= -8+4/6= -4/6= -2/3.

X= -8-4/6= -12/6= -2.

Answer is: -2/3 and -2.

You might be interested in

Which of the following is a solution of x^2 + 2x + 4?

Kaylis [27]

Answer:

$\displaystyle x_1=-1+\sqrt{3}i$

$\displaystyle x_2=-1-\sqrt{3}i$

Step-by-step explanation:

<u>Second-Degree Equation</u>

The second-degree equation or quadratic equation has the general form

$ax^2+bx+c=0$

where a is non-zero.

There are many methods to solve the equation, one of the most-used is by using the solver formula:

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The equation of the question has the values: a=1, b=2, c=4, thus the values of x are

$\displaystyle x=\frac{-2\pm \sqrt{2^2-4\cdot 1\cdot 4}}{2\cdot 1}$

$\displaystyle x=\frac{-2\pm \sqrt{-12}}{2}$

Since the square root has a negative argument, both solutions for x are imaginary or complex. Simplifying the radical

$\displaystyle x=\frac{-2\pm 2\sqrt{-3}}{2}=-1\pm\sqrt{3}i$

The solutions are

$\displaystyle x_1=-1+\sqrt{3}i$

$\displaystyle x_2=-1-\sqrt{3}i$

7 0

3 years ago

Please help (no links or I will report you)

d1i1m1o1n [39]

Answer:

0.60 or 0.6

Step-by-step explanation:

Divide three by five :)

3÷5= 0.6

8 0

3 years ago

Is f(x) = 10x2 + 10, x > 0 the inverse of g(x) = <img src="https://tex.z-dn.net/?f=%5Csqrt%20%5Cfrac%7Bx-10%7D%7B10%7D" id="T

agasfer [191]

Step-by-step explanation:

$x > - \frac{ 1}{5}$

6 0

3 years ago

Classify each polynomial according to its degree and type. Look at the screenshot below!

Ahat [919]

Answer:

Monomial Binominal Trimoninal

Degree 1: -9x x + 6

Degree 2: -4 $x^{2}$ $4x^{2} -x$ $x -3x^{2} + 1$

Degree 3: 4 $x^{3}$ 5 - 2 $x^{3}$ $3x^{2} + 3x^{3} -10$

Step-by-step explanation:

I do not know how to explain how I got to this answer, but here is the answer.

8 0

2 years ago

How many integers between 100 and 999 (inclusive) have distinct digits

daser333 [38]

Answer:

648

Step-by-step explanation:

Running this in Python, with the code as follows,

import math

cur_numbers = [0] * 3

num = 0

for i in range(100, 1000):

cur_numbers[2] = i % 10

i = math.floor(i/10)

cur_numbers[1] = i % 10

i = math.floor(i/10)

cur_numbers[0] = i % 10

if(len(set(cur_numbers)) == 3):

num += 1

print(cur_numbers)

print(num), we get 648 as our answer.

Another way to solve this is as follows:

There are 9 possibilities for the hundreds digit (1-9). Then, there are 10 possibilities for the tens digit, but we subtract 1 because it can't be the 1 same digit as the hundreds digit. For the ones digit, there are 10 possibilities, but we subtract 1 because it can't be the same as the hundreds digit and another 1 because it can't be the same as the tens digit. Multiplying these out, we have

9 possibilities for the hundreds digit x 9 possibilities for the tens digit x 8 possibilities for the ones digit = 648

5 0

3 years ago