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Find the intercepts given the equation 5x + 2y = 10.
Let me see.
To find the x-intercept let y = 0.
5x + 2y = 10
5x + 2(0) = 10
5x + 0 = 10
5x = 10 - 0
5x = 10
x = 10/5
x = 2
To find the y-intercept, let x = 0.
5x + 2y = 10
5(0) + 2y = 10
0 + 2y = 10
2y = 10 - 0
2y = 10
y = 10/2
y = 5
The x-intercept is x = 5, which can also be written (5, 0).
The y-intercept is y = 5, which can be also be written (0, 5).
You say?
The midpoint of the line segment from A to B is (5, -4). If B = (7, -2), what is A?
Solution:
Let point A = (x_1, y_1).
Here is my set up to find x_1.
5 = (x_1 + 7)/2
Here is my set up to find y_1.
-4 = (y_1 - 2)/2
Is this the correct set up to find the coordinates of point A?
This often happens in maths. You suspect a general result is true. Step 1**: Try out some particular values to see if it's true in those cases. Step 2: Then try to show it's always true by substituting letters for numbers.
I've shown that it's true for root 7. Now to test in general. So choose a letter to replace 7. At first I chose x but then I got into trouble using x to mean times as well. So I switched to a different letter. I chose 'r' to stand for 'root'.
Bob
** Strictly if you can do step 2 you don't need step 1, as step 2 shows it's always true. But, if you're struggling to understand a topic, trying number values first helps.
Thanks for clearing that up.
harpazo1965 wrote:The area of a rectangular window is to be 306 square centimeters.
If the length exceeds the width by 1 centimeters, what are the dimensions of the window?I think A = length • width is needed here. Yes?
Length = x + 1
Width = x
306 = (x + 1)(x)
(x + 1)(x) = 306
x^2 + x = 306
x^2 + x - 306 = 0
Is this the correct equation?
solve the eqn and check the sol'ns in the problem
you'lla get 2 sol'ns
ignore the negative sol'n
After solving the quadratic equation for x, I got two answers like you said.
x = -18
x = 17
We are talking about distance. So, I reject -18.
Answer: width = 17 centimeters; length = 18 centimeters.
yes
Ok. Cool. I will work it out on paper.
harpazo1965 wrote:Factor 2.3x^2 + 1.4x - 4.5 = 0 without a calculator.
What is the best method to use for factoring equations with decimals without a calculator?
i'd multiply thru by 10 to get rid of the decimal pts
then do the Quadratic Formula
answer is messy
This is exactly what I thought about doing at first. I don't mind messy answers as long as it is right.
harpazo1965 wrote:Given (π/2)^2 = π^2/4, I now must add π^2/4 to both sides of the equation.
Here it is:
x^2 + (π/2) x + π^2/4 = -1 + π^2/4
Where do I go from here?
turn the left-hand side into perfect-square form
combine the right-hand side to being 1 fraction
take the square root of both sides
solve for x= on the left-hand side
x^2 + (π/2) x + π^2/4 = -1 + π^2/4
(x + (π/2))(x + (π/2)) = 1.4674
(x + (π/2))^2 = 1.4674
I now take the square root on both sides. Yes?
harpazo1965 wrote:Plotting points tells me where the intercepts are located. Yes? Do you have an example graph for me?
make your own example (& look at the examples in the book)
like take (x - 2)(x + 3) = y
graph it & check that it crosses at x = 2, x = -3
Copy. Will do.
Is a number (A) the square root of another number (B) ?
The only requirement is that A^2 = B.
As -√ 7 x -√ 7 = 7 it is a square root of 7.
If √r is a square root of y, then -√r times -√r = y as multiplying two minuses makes a plus.
So this happens for every positive number.
Negative numbers don't have a real number root. (But see complex numbers for more on this)
Bob
In your reply, where did sqrt{r} come from?
Where did the variable [ r ]come from?
Bob wrote:This business of rounding part way through a problem can lead to very inaccurate answers. here's an example:
Given x^2 = 56 work out x^5
On a calculator x = 7.483314774...
If I call that 7.5 and work out x^5 I get 23730.46875
If I use the full value I get 23467.67513.....
You can see that using the rounded up number and multiplying five times leads to an answer that is quite a bit too big.
Yes! Amen! Wait to round until the end!
I get it now. I found this problem to be uniquely interesting.
Yes, that's right.
At this point
35.3553 = s
So, the side of the square is about 35 feet.
you make an approximation and then use this to continue.
At the end you make another to reach the answer.
It's better to avoid approximations like this:
Say s = √ 1250 and leave it like this.
Then c^2 = ( √ 1250 )^2 + ( √ 1250 )^2 = 1250 + 1250 exactly. So there's no need to make the two approximations. The answer is exactly 25.
Your method is OK but I used a quicker method that led to 25 directly:
Sketch the square and draw its diagonals, splitting the square into four triangles. The distance from the centre of the square to any vertex is the radius we want so call it x.
area of a triangle = half base times height = 0.5 times x^2
Four triangles makes the square so we have the equation
If you want to really 'show off' you can say that -25 also fits the equation but is not an answer as a positive answer is needed.
This business of rounding part way through a problem can lead to very inaccurate answers. here's an example:
Given x^2 = 56 work out x^5
On a calculator x = 7.483314774...
If I call that 7.5 and work out x^5 I get 23730.46875
If I use the full value I get 23467.67513.....
You can see that using the rounded up number and multiplying five times leads to an answer that is quite a bit too big.
Bob
I didn't think I would get the right answer. I simply tried to do my best work. Your method leading to 25 feet is pretty cool.
An adjustable water sprinkler that sprays water in a circular pattern is placed at the center of a square field whose area is 1250 square feet. What is the shortest radius setting that can be used if the field is to be completely enclosed within the circle?
I will need the area of a square formula: A = (side)^2.
I know that A = 1250 (feet)^2.
1250 = s^2
sqrt{1250} = sqrt{s^2}
35.3553 = s
So, the side of the square is about 35 feet.
I need the hypotenuse of the square.
a^2 + b^2 = c^2
(35)^2 + (35)^2 = c^2
1225 + 1225 = c^2
2450 = c^2
sqrt{2450} = sqrt{c^2}
49.49747468306 = c
Let c = hypotenuse = diagonal of square.
To find the radius distance, I divide c by 2.
So, c/2 = 24.7487. If I round this decimal number to the nearest first place, I get 25 feet.
I say the shortest radius setting is about 25 feet.
You say?
The area of a rectangular window is to be 306 square centimeters.
If the length exceeds the width by 1 centimeters, what are the dimensions of the window?
I think A = length • width is needed here. Yes?
Length = x + 1
Width = x
306 = (x + 1)(x)
(x + 1)(x) = 306
x^2 + x = 306
x^2 + x - 306 = 0
Is this the correct equation?
How many right triangles have a hypotenuse that measures
2x + 3 meters and legs that measure 2x - 5 meters and x + 7 meters?
What are the dimensions of the triangle(s)?
I think a^2 + b^2 = c^2 cane be used here.
Let a = 2x - 5
Let b = x + 7
Let c = 2x + 3
Is the correct set up as shown below?
(2x - 5)^2 + (x + 7)^2 = (2x + 3)^2
Must I now solve for x?
2.3x^2 + 1.4x - 4.5 = 0
Wolfram Alpha doesn't get an easy factorisation here. Are you sure that's copied correctly. But WA does do what I'd do anyway which is to remove the 2.3 as a 'common' factor:
You can then complete the square as shown elsewhere, but you don't get rational coefficients so use of a calculator is essential hence why I question the posted quadratic.
Bob
Maybe it is a typo at my end. What about if I clear the decimal numbers multiplying by a whole number(s) that would allow me to do so?
Example:
Say the coefficient of the x^2 term is 0.10.
I can multiply 0.10 by 100 to get 10. Can this be done to the decimal
numbers given in this question (considering it is not a typo)?
No because when multiplied out you'll get an x^4 term.
3x^3 - x^2 +18x - 6
Bob
I often get the wrong answer when rushing through basic problems.
Every positive number (except 0) has two square roots, one positive and one negative, so you need to allow for both possibilities. Sometimes the context of the question will then rule out one answer.
x cannot be both -5 and 9 at the same time so it has to be or
Bob
Why does every positive number have two square roots?
Take 7, for example. Are you saying that 7 has both -sqrt{7} and sqrt{7}?
Why is this the case?
hi
That's a bit confused.
You're right to make the x^2 coefficient 1. Then you need to add something to make the correct constant term.
Take a look at a perfect square:
So if your quadratic looks like this:
.....(1)you need to add a^2
Here's an example:
Comparing with (1) I can see that a=2
a^2 = 4 so that's what I need to add (12 to each side)
The left hand side is a perfect square so I can make this factorisation
Bob
I added to both sides in my question. Do you see it?
Look here:
Given (π/2)^2 = π^2/4, I now must add π^2/4 to both sides of the equation.
Here it is:
x^2 + (π/2) x + π^2/4 = -1 + π^2/4
Where do I go from here?
Factor 2.3x^2 + 1.4x - 4.5 = 0 without a calculator.
What is the best method to use for factoring equations with decimals without a calculator?
You say?
Factor 3x^3 - x^2 +18x - 6 by grouping.
I will divide the four terms into two terms.
3x^3 - x^2...Group A
18x - 6...Group B
I will factor each group individually.
Group A factors out to be x^2(3x^2 - 1).
I see that 3x^2 - 1 can also be factored to be (3x - 1)(x + 1).
Group B factors out to be 6(3x - 1).
Putting it a together I get this:
x^2(3x - 1)(x + 1)6(3x - 1)
I think the final answer looks like this:
6x^2(3x - 1)(x + 1)
Is this right?
Solve ( x − 2)^2 = 49 using the Square Root Method.
Let me see.
I know that taking the square root on both sides of the equation is step 1.
sqrt{( x − 2)^2} = sqrt{49}
x - 2 = -7 or x - 2 = 7
x = -7 + 2 or x = 7 + 2
x = -5 or x = 9
Question:
Why do we get a negative answer and a positive answer when taking the square root?
Questions:
Do I use or? Do I use and?
Take the answers here.
Do I write the answers as
x = -5 or x = 9?
Do I write the answer as
x = -5 and x = 9?
You say?
Solve equation by completing the Square.
2x^2 + π x + 2 = 0
Let me see.
In order to complete the square, the coefficient of the x square term
cannot be greater than or less than 1.
So, I must divide each term on both sides of the equation, in this case, by 2.
(2x^2 + π x + 2)/2 = 0/2
Doing so, I get the following:
x^2 + (π/2) x + 1 = 0
I now subtract 1 from both sides.
x^2 + (π/2) x = -1
The next step is to raise the coefficient of x to the second power and add the product to both sides of the equation.
(π/2)^2 = π^2/4
x^2 + (π/2) x + π^2/4 = -1 + π^2/4
Stuck here....
Tell them you'll look after them when they develop dementia.
Bob
Good point. You are right. Keeping our memory power active is so important as we age.
Plot some points, and have a look.
Bob
Plotting points tells me where the intercepts are located. Yes? Do you have an example graph for me?
Good to me.
B
My friends and family think I am crazy for solving math problems at 58 years old. They ask: HEY, WHAT'S THE POINT?