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3 years ago

PLEASE HELP I DONT UNDERSTAND

Mathematics

1 answer:

Charra [1.4K]3 years ago

8 0

Example Question:

X²+6x-14=0

Quadratic equations are usually in the form Ax²+Bx=C. Completing the Square is a technique for rewriting this equation in the form (x+a²)+b.

The example I provided is in standard form.

Here's what I did:

Step 1: Move the "C" to the other side (in this case it's negative 14, so you add 14 to both sides of the equation to move it).

Step 2: Leave a blank after Ax²+Bx. In this case, it'll look like: X²+6x+_____= 14.

Step 3: Use the box to find the missing "C". Idk if you can see the photo after the 1st one, but I circled the spaces in green. Those spaces indicate 2 factors that add up for B (in this case, 6). Those same factors will eventually multiply each other to get 9.

Step 4: Give the new "C" to both sides of the equation. What this means is that after you multiply the factors on the outside of the box, you put the number on both sides. The 3rd picture I drew is the process of finding the outer factors (by plugging in 3x to the spaces that were circled).

If you can't see it, it's (X+3)•(X+3).

The 4th picture is me plugging the new "C" (9) to both sides of the equation. 3 times 3 is 9. You multiply the 3's on the outside to get the factors on the inside.

Step 5: Use Step 3 box to write (perfect square)² of left hand side.

What this means is that you take the factors on the outside and rewrite it in perfect square form. like shown in picture 5. If you can't see it, it's like this:

(X+3)•(X+3) turns to (X+3)²=23.

The 23 was from adding 14+9 together.

So the final Answer would be:

(X+3)²=23.

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What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

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Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$