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3 years ago

How do you solve a quadratic equation by factoring

Mathematics

2 answers:

Darya [45]3 years ago

6 0

Answer:

Step-by-step explanation:

Step 1: Obtain zero on one side and then factor. Step 2: Set each factor equal to zero. Step 3: Solve each of the resulting equations. This technique requires the zero factor property to work so make sure the quadratic is set equal to zero before factoring in step 1.

kow [346]3 years ago

5 0

If were you, i will use trial and error to make it easier :)) [if the number is small,you can use trial and error, but if a big number use quadratic eq. Or divide the big num. to get smaller num.]

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Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

3 years ago

Use the grouping method to factor this polynomial completely. 3x3 + 12x2 + 2x + 8

dezoksy [38]

Answer:

(3x^2 + 2)(x + 4).

Step-by-step explanation:

3x3 + 12x2 + 2x + 8

3x^2(x + 4) + 2(x + 4) The x + 4 is common to the 2 groups so we have:

(3x^2 + 2)(x + 4).

3 0

3 years ago

What is the factor of x²-4x+24

DochEvi [55]

Answer:

Their are no possible factors for this expression.

Step-by-step explanation:

3 0

3 years ago

Some pls help me I’ll give out brainliest please dont answer if you don’t know

Goshia [24]

Answer:

move the constant to the right hand of the side and change the sign

2x>-0.5-5-5

calculate the difference

2x>-6

divide both sides of the inequality by 2

x>-3

solution X>-3

3 0

3 years ago

What is an equation of a circle with center (2,7) and radius 4?

goblinko [34]

Equation of a circle:

(x - h)^2 + (y - k)^2 = r^2

In this case:

center (2,7) and radius 4 so h = 2, k = 7 and r = 4

Equation:

(x - 2)^2 + (y - 7)^2 = 4^2

(x - 2)^2 + (y - 7)^2 = 16

Hope it helps.

8 0

3 years ago