Find their respective speeds

this is how i simplified the equation in post#93

40^2=4x2+ 4x2 +32x+64

1600= 8x2 +32x+64 (I have to turn this now into a quadratic equation to be able to solve it)

8x2 +32x+64-1600=0

8x2 +32x – 1600=0

Is it good?. Can I go on to apply the quadratic formula to solve this equation?. Or is there something wrong?. Please, confirm.
 
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Sorry, I missed a step. I did not do the subtraction.
8x2 +32x+64-1600=0
8x2 +32x – 1536=0 ....... be careful!!
 
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eddy2017 said:
this is how i simplified the equation in post#93

40^2=4x2+ 4x2 +32x+64

1600= 8x2 +32x+64 (I have to turn this now into a quadratic equation to be able to solve it)

8x2 +32x+64-1600=0

8x2 +32x – 1536=0

Is it good?. Can I go on to apply the quadratic formula to solve this equation?. Or is there something wrong?. Please, confirm.
Click to expand...
I would go one more step:

8x2 +32x – 1536=0 ....................... Divide by 8

x2 +4x – 192 = 0

Now apply the formula....
 
eddy2017 said:
Sorry, I missed a step.
8x2 +32 – 1536=0
Click to expand...
The last step is incorrect. You lost x from 32x and 1600 somehow became 1536.
You can apply the quadratic formula to 8x2 +32x – 1600=0 but it can be further simplified to make things easier. You can divide both sides by 8.
 
lev888 said:
The last step is incorrect. You lost x from 32x and 1600 somehow became 1536.
You can apply the quadratic formula to 8x2 +32x – 1600=0 but it can be further simplified to make things easier. You can divide both sides by 8.
Click to expand...
1536 is correct because:

1600 - 64 = 1536
 
lev888 said:
The last step is incorrect. You lost x from 32x and 1600 somehow became 1536.
You can apply the quadratic formula to 8x2 +32x – 1600=0 but it can be further simplified to make things easier. You can divide both sides by 8.
Click to expand...
Thanks, got it. Because 8 is a the greatest common factor between 8,32,and 1600?.
 
eddy2017 said:
this is how i simplified the equation in post#93

40^2=4x2+ 4x2 +32x+64

1600= 8x2 +32x+64 (I have to turn this now into a quadratic equation to be able to solve it)

8x2 +32x+64-1600=0

8x2 +32x – 1600=0

Is it good?. Can I go on to apply the quadratic formula to solve this equation?. Or is there something wrong?. Please, confirm.
Click to expand...
As you noted, you messed up on the subtraction.

As Subhotosh suggested, dividing through by 8 simplifies future arithmetic.

And why do you ask whether applying the quadratic is OK, or were you asking for confirmation that you had made no errors before you bothered with the formula?
 
Okay. I gave it my best try. Thanks for your advice and hints and plenty of hand-holding. I am so thankful!
I wrote it down on a word doc. I will see if the results paste well here, or, i will attach the page.


1600= 8x2 +32x+64 (I have to turn this now into a quadratic equation to be able to solve it)

8x2 +32x+64-1600=0

8x2 +32x – 1536=0 (now you can simplify by dividing both sides by 8 the g.c.f



8x2 +32x – 1536= 0

8 8



x^2+4x-192=0



x= -b ± √b2-4ac

2a

I have to find the values for a, b and c in the quadratic equation

x^2+4x-192=0

a=1

b=4

c= -192

now that I have identified these values, I’m going to plug them in our formula

x= -b ± √b2-4ac

2a



x= -4 ± √162-4(1*-192)

2

x= -4 ± √16 -4(-192)

2

x= -4 ± √16+768

2

x= -4 ± √16+768

2

x= -4 ± √784

2

x= -4 ± 28

2

x1= -4 + 28

2

x1=24 =12 km/h car 1 has a speed of 12 km/h with the positive vale +)

12

Let’s try the other symbol –

x2= -4 ± 28

2



x2= -4 - 28

2



x2= -32

2

-16 km/p which is not a valid solution



So, car 1 rate of speed =12 kmph

Car 2 rate of speed is 12 + 4 =16 kmph
 
When I save the text does not look good. Numbers scoot to the left. If you know i can avoid this, please, let me know.
 
You cannot (without tremendous effort) type fractions that way on this site unless you know LaTeX. Use slash and parentheses.

You did the quadratic formula correctly. WELL DONE. But you failed to do the last step, CHECK YOUR WORK. The way to do that is to see whether the distances work out.

2 hours at 12 miles per hour is a distance of 24 miles.

2 hours at 16 miles per hour is a distance of 32 miles.

[MATH]\sqrt{24^2 + 32^2} = \sqrt{576 + 1024} = \sqrt{1600} = 40.\ \checkmark[/MATH]
 
Wow!!. Thanks a lot.
Thanks a lot to all of you!.
Already looking forward to my next challenge.!
Thanks.
 
eddy2017 said:
When I save the text does not look good. Numbers scoot to the left. If you know i can avoid this, please, let me know.
Click to expand...
Here is an example of the easy way to type fractions on here. This is the quadratic formula, written mostly as we would with no special formatting system, and with careful usage of parentheses:

x=(-b pm sqrt(b^2-4ac))/(2a)​

When I put that between backquotes, the system formats it nicely:

`x=(-b pm sqrt(b^2-4ac))/(2a)`​
 
T
Dr.Peterson said:
Here is an example of the easy way to type fractions on here. This is the quadratic formula, written mostly as we would with no special formatting system, and with careful usage of parentheses:

x=(-b pm sqrt(b^2-4ac))/(2a)​

When I put that between backquotes, the system formats it nicely:

`x=(-b pm sqrt(b^2-4ac))/(2a)`​
Click to expand...
Thanks a lot. I'll try that.
 
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