Solving Algebraic Equation: Find value of k = 4 - b^2 + ac for a = 0.25, b = -2.15, c = -0.9

Scooterat

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Hi everyone first post on here, having trouble solving this equation, as I'm getting a different answer from a equation calculator :(



Find the value of [math]k\, =\, 4\, -\, b^2\, +\, ac[/math] when [math]a\, =\, 0.25,\, b\, =\, -2.15,[/math] and [math]c\, =\, -0.9.[/math] Give your answer correct to two significant figures.



k = 4 - B2 + ac

K - AC= 4 - B2

K - AC + B2 = 4

4 = K - AC + B2

4 = K - AC + (-4.62225)

4 = K + 0.65 + (-4.6225)

4 = K - 3.97225
 

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They aren't asking you to solve an equation. Just plug the numbers into the expression and evaluate it:

[MATH]k = 4 - (-2.15)^2 + (0.25)(-0.9) =[/MATH] ...​
 
Tell me why you chose to retain only two decimal places.
 
You are making this way too difficult.

[MATH]k= 4 - b^2 + ac[/MATH]
[MATH]b= -\ 2.15 \implies b^2 = 4.6225 \implies -\ b^2 = -\ 4.6225 [/MATH]
[MATH]a = 0.25 \text { and } c = -\ 0.9 \implies ac =-\ 0.225[/MATH]
[MATH]\therefore k = 4 - 4.6225 - 0.225 = 4 - 4.8475 = -\ 0.8475 \approx -\ 0.85.[/MATH]
All your algebra was unnecessary, nor did you complete it. If you had completed it, you would have got the correct answer except that you had made an error in arithmetic.

Note that [MATH]0.25 * -\ 0.9 = -\ \dfrac{1}{4} * \dfrac{9}{10} = -\ \dfrac{25}{100} * \dfrac{9}{10} = \dfrac{225}{1000} = -\ 0.225 \ne 0.65.[/MATH]
 
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Correct....-0.8475 to be precise.

Any more questions?!
 
You're saying it was just accidental or your calculator was set to 2? I was hoping you would say something about "significant figures".

Lesson learned on this one. No need to "solve" if that's not what you were asked to do.

It was just a mistake on my part, I'm having to take a maths unit as part of my university course and I'm very rusty, again thankyou for your help and I will most likely be posting more in the future. I had to do some work on significant figures earlier on the unit. Thankyou for your help Tk :)
 
You are making this way too difficult.

[MATH]k= 4 - b^2 + ac[/MATH]
[MATH]b= -\ 2.15 \implies b^2 = 4.6225 \implies -\ b^2 = -\ 4.6225 [/MATH]
[MATH]a = 0.25 \text { and } c = -\ 0.9 \implies ac =-\ 0.225[/MATH]
[MATH]\therefore k = 4 - 4.6225 - 0.225 = 4 - 4.8475 = -\ 0.8475 \approx -\ 0.85.[/MATH]
All your algebra was unnecessary, nor did you complete it. If you had completed it, you would have got the correct answer except that you had made an error in arithmetic.

Note that [MATH]0.25 * -\ 0.9 = -\ \dfrac{1}{4} * \dfrac{9}{10} = -\ \dfrac{25}{100} * \dfrac{9}{10} = \dfrac{225}{1000} = -\ 0.225 \ne 0.65.[/MATH]

My tutor has asked for it to be laid out in this format with each single step being noted. Very rusty as I haven't studied maths for two years, Thankyou for your input
 
To reassure you, the reason you wrote only two decimal places was simply that the problem said to! It said, "Give your answer correct to two significant figures," which in this case is the same as two decimal places (because all digits are decimal places). It was not a mistake.
 
I think the exercise statement could be better worded. (They began by asking for the value of an equation.)

Also, how can we report the answer to two significant digits when 0.9 shows only one significant digit?

?
 
I think the exercise statement could be better worded. (They began by asking for the value of an equation.)

Also, how can we report the answer to two significant digits when 0.9 shows only one significant digit?

?

I read it as asking for the value of k, which is equal to ... . I think it's a fairly common idiom: technically wrong, but understood.

And it doesn't say to give the answer in the appropriate number of significant digits, based on the data; it just says to write two significant digits. Maybe they know that 0.9 is exact. Whatever the reason, you can do what you are told. Yes, a person who is very familiar with significant digits will cringe as he does it, but I suspect that the context is not one in which significant digits are a primary emphasis. (I say that, of course, because if it were, they would be wrong -- not because I know any more than we have been told.)
 
… I suspect that the context is not one in which significant digits are a primary emphasis …
Obviously! (They could have just said, "rounded to two decimal places", if that's what they want.)

I mention the error in reporting two significant digits (based on the numbers as given) for the benefit of future readers who may be confused because they're looking for information about significant digits.

I think the exercise could be better worded.

?
 
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