Math help

[Chaos_Sorcerer]

It's the summer, so I don't technically have school, but I like to solve math problems every now and then, to sort of keep myself in mental shape. This latest one stumped me. Attached is the full problem and solution.

 

POTWE-16-AE-GM-23-S.pdf

 

I get most of the solution, but can someone explain to me how to arrive at 8 m being the radius without random guessing? Basic simplification got me to x^2 + x − 72 = 0, and then I see how you can work backwards to get to (x − 8)(x + 9) = 0, but this was a weird way to me, and I don't understand how they determined that x = 8 or - 9 from it. I simplified it to x^2 = 72 - x, and then determined that it was 8 by just guessing. Looking at it now, I see how x can also be - 9 (just by plugging it into x^2 = 72 - x).

 

I've only really learned grade 9 math, so I may not be familiar with concepts taught in higher education. (Just a side note so no one assumes that I know a bunch of stuff I don't.)

[Dash Lambda]

The quantity you're solving for is a length. Since lengths can't be negative, the negative solution is irrelevant, leaving only the positive solution x=8.

[Chaos_Sorcerer]

Just now, Dash Lambda said:

The quantity you're solving for is a length. Since lengths can't be negative, the negative solution is irrelevant, leaving only the positive solution x=8.

But how did you get 8 and - 9 in the first place?

[Shakaza]

12 minutes ago, Chaos_Sorcerer said:

But how did you get 8 and - 9 in the first place?

Basically, like this: (8 - 8)(-9 + 9) = 0 is equivalent to (0)(0) = 0. You insert the value of x that would cause the left side to be equal to the right side, and with the simplification of that equation, the right side is always 0. And, since both parts on the left side are being multiplied together, having just one of them equal to 0 will result in that whole side equaling 0, hence why x could be either value. If you do, this, for example: (8 - 8)(8 - 9) = 0. Same result if you insert the other number.

[Chaos_Sorcerer]

2 minutes ago, Shakaza said:

Basically, like this: (8 - 8)(-9 + 9) = 0 is equivalent to (0)(0) = 0. You insert the value of x that would cause the left side to be equal to the right side, and with the simplification of that equation, the right side is always 0. And, since both parts on the left side are being multiplied together, having just one of them equal to 0 will result in that whole side equaling 0, hence why x could be either value. If you do, this, for example: (8 - 8)(8 - 9) = 0. Same result if you insert the other number.

Didn't get what you were saying at first, but it makes so much sense now. Thanks!

 

Is there a decent guide on how to use distributive property backwards?

[Dash Lambda]

29 minutes ago, Chaos_Sorcerer said:

Didn't get what you were saying at first, but it makes so much sense now. Thanks!

 

Is there a decent guide on how to use distributive property backwards?

Solving equations by factoring (what you're talking about) is a fundamental topic in Algebra -The best way to learn it is to go through Algebra.

The reason I say that is because it's actually a pretty important and varied building block -It's not like using the quadratic equation where you simply learn where to apply it to what, it involves manipulation of equations to fit certain patterns.

 

Though, for quadratic equations, it's pretty rigid: Given an equation x²+bx+c, with the factored form (x+p)(x+q) which expands to x²+(p+q)x+pqx, you know that (p+q)=b and pq=c, and you simply solve the system. You can then find the solution as @Shakaza explained.

[ionbasa]

A slightly different approach:

Once you reach x^2 + x +72 , Instead of factoring, you can use the quadratic formula:

In the equation you get: a = 1, b =1, and c = 72 (a,b,c are the coefficients from your equation). From there you can simplify to get x = 8 and x = -9. We know that lengths cannot be negative, so in this case, we are only concerned with the positive result (8).

 

Factoring can also only get you so far. Sometimes there are tricks we can do if we have a polynomial of a higher degree ( x^3, X^4, etc...). Factoring is an important part of algebra, and its handy to know, the vast majority of factoring involves recognizing patters. Try practicing foil/distributing first and work backward with the same problem. 

[Chaos_Sorcerer]

@Shakaza @Dash Lambda @ionbasa

 

Thank you all for your input.

Out of curiosity...is this taught in the 9th grade in the US? 

[ionbasa]

43 minutes ago, Chaos_Sorcerer said:

@Shakaza @Dash Lambda @ionbasa

 

Thank you all for your input.

Out of curiosity...is this taught in the 9th grade in the US? 

It varies, but Algebra 1 is introduced to students in 8th grade, where student learn the fundamentals, such as factoring and the quadratic equation. 9th grade is intro to Trigonometry. 10th grade students tend to learn Algebra 2 , which builds up on both Trigonometry and Algebra 1.

[Chaos_Sorcerer]

1 hour ago, ionbasa said:

It varies, but Algebra 1 is introduced to students in 8th grade, where student learn the fundamentals, such as factoring and the quadratic equation. 9th grade is intro to Trigonometry. 10th grade students tend to learn Algebra 2 , which builds up on both Trigonometry and Algebra 1.

So us Canadians are a little behind, then. Problem E is apparently for students in the 11th and 12th grades (the one attached)...maybe we need to revamp our education system...