All categories

3 years ago

What is the first step in solving a quadratic equation of the form given below? (ax + b)2 = c

Mathematics

2 answers:

Aleonysh [2.5K]3 years ago

8 0

Multiply the 2 into the ax and b, and then moving the c over and changing it's sign

hope this helps

Alexxx [7]3 years ago

4 0

Answer:

The answer on apex is: Take the Square Root of both sides.

Step-by-step explanation:

You might be interested in

What is the sum of 5x^2y and (2xy^2 + x^2y)

eimsori [14]

Recognize that you must combine "like" terms. 5x^2y and x^2y are "like" terms.

Adding them together, you get 6x^2y.

Now add 2xy^2 to 6x^2y. These are NOT "like" terms, so you end up with
6x^2y+2xy^2 as your final answer. This answer is acceptable as is.

However, you could factor out the common factors: 2xy(3x+y).

8 0

2 years ago

What would be the value of X

Alex_Xolod [135]

To find the value of x, you need to set up a proportion.This is the proportion you would set up:

2.5 x
---- = ------
3.75 60

x=40

7 0

3 years ago

Read 2 more answers

Calculus piecewise function.

Kipish [7]

Part A

The notation $\lim_{x \to 2^{+}}f(x)$ means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc

We don't actually arrive at x = 2 itself. We simply move closer and closer.

Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2

Therefore f(x) = (x/2) + 1

Plug in x = 2 to find that...

f(x) = (x/2) + 1

f(2) = (2/2) + 1

f(2) = 2

This shows $\lim_{x \to 2^{+}}f(x) = 2$

Then for the left hand limit $\lim_{x \to 2^{-}}f(x)$ , we'll involve x < 2 and we go for the first piece. So,

f(x) = 3-x

f(2) = 3-2

f(2) = 1

Therefore, $\lim_{x \to 2^{-}}f(x) = 1$

===============================================================

Part B

Because $\lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x)$ this means that the limit $\lim_{x \to 2}f(x)$ does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

===============================================================

Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

===============================================================

Part D

As mentioned earlier, since $\lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3$ , this means the limit $\lim_{x \to 4}f(x)$ does exist and it's equal to 3.