Math Equation

I'm having trouble with a couple of math equations here I was given to solve.

Here is my problem Solve the quation the square root of 3x - 8 minus the square root of x-4 equals 2.

and here is an example of my problem.

the square root of 2x minus 1 minus the square root of x-5 equals 3.


Isolate one of the radicals: the square root of 2x - 1 = the square root of x - 5 + 3

Square both sides: 2x-1 = x-5+6 the square root of x-5 + 9


This is as far as I got when I was trying to apply my problem to this one. I'm confused where this six came from. I don't think this program is showing all the steps.




the sum of three consecutive positive even intergers is equal to two more than seven times half the number. What are the three even positive intergers?

To me I'm imagining it looks like this.

a + b + c = 2 + 7 x 1/2n what's a,b, and c?

It's a little more confusing to me with the even numbers portion. To me just the 2 and the seven doesn't seem like enough information to solve the problem.

All help will be greatly appreciated.

Notation: x^1/2 is also the same as a square root. Do not be misled. It means the same thing as something being under a radical.

So ultimately, what we want to do here, is that we want to find a value for the variable x. Unfortunately, it's hidden underneath of a square root. Though that might normally be trivial, we can assume that we should not just plug stuff into the calculator and be done with it. This means we have to get through the algebra. We'll need our FOIL methods, and you'll see why soon enough.

Let's write our equation; I assume it looks like this, but if it's wrong, try to correct me:
(3x - 8)^1/2 - (x - 4)^1/2 = 2

There's no way to square one side without still getting a radical somewhere. So, bring one of the terms over to the other side; in this case, we'll just use the second term, to make things simple:
(3x - 8)^1/2 = 2 + (x - 4)^1/2

Square both sides:
((3x - 8)^1/2)² = (2 + (x - 4)^1/2)²

Simplify; we need our foiling skills here:
(3x - 8) = 4 + 2(x - 4)^1/2 + x - 4

Cancel out the 4; isolate the radical by bringing the 2 and the x to the other side
(3x - 8 - x)/2 = (x - 4)^1/2

Combine like terms (as in, the variable x) and then square again:
((2x - 8)/2)²=((x - 4)^1/2)²

Simplify the 2 in the denominator, and clean up:
(x - 4)² = (x - 4)

Expand our squared term, and bring the right side over:
x² - 8x + 16 - x + 4 = 0

Combine like terms:
x² - 9x + 20 = 0

This looks like a regular polynomial. Factors of 20 are 1, 20; 2, 10; 4, 5. One of those pairs adds up to nine. So, factor terms:
(x - 5)(x - 4) = 0

Solve for x so that the equation equals zero:
x = 4 or 5

Does this make sense? Yes. Substituting in 4 or 5 into our original equation would not get any non-real answers. Additionally, since our highest exponent became a term to the second power, we should reasonably obtain two answers. Both 4 and 5 are separate and distinct answers.

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This is another case of foiling gone horribly imbasauce. What else is there to do other than to factor stuff out?

Okay, write your givens:
x⁴ - 12x² - 85 = 0.

Factors of 85 are indeed, 17 and 5. However, that twelve there is negative, which is 17 - 5. The 85 is also negative, which means that both of the terms in the parenthesis will not necessarily be the same sign.
(x² - 17)(x² + 5) = 0

Here's where you went wrong: the solution to (x² + 5) = 0 is not +/- (5)^1/2.
x² = -5.

The solution is +/- (5i)^1/2. i, being the square root of -1, when squared, will give us a negative, which is what is needed to solve the problem. So, our new solutions are:
+5i^1/2
-5i^1/2
+17^1/2
-17^1/2

+/- (17)^1/2 remains unchanged. Your initial guess about the solution to the second equation was correct.

EDIT: Okay, so you changed the situation on me. In this case, your guess with +/- (17i)^1/2 is incorrect. Though +/- (5i)^1/2 is correct, that does not mean that it applied to the 17, which had a negative factor with your x². That makes it normal. The positive factor in front of your 5 with x², made it need i as part of your solution.

Quote:

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(3x - 8) = 4 + 2(x - 4)1/2 + x - 4

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This is where you squared right? I understand where you got that first 4, but i'm unsure about the 2 and the x-4.

In comparison to the example problem at this step. I don't know how to use subscripts like you so I apologize.

2x-1 = x-5+6 the square root of x-5 + 9


Just to mention in the hinto I'm shown there is no paretheses around the square root of x-5 is it still assumed to multiply?

Quote:

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Originally Posted by Depression Moon

Quote:

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(3x - 8) = 4 + 2(x - 4)1/2 + x - 4

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This is where you squared right? I understand where you got that first 4, but i'm unsure about the 2 and the x-4.

In comparison to the example problem at this step. I don't know how to use subscripts like you so I apologize.

2x-1 = x-5+6 the square root of x-5 + 9


Just to mention in the hinto I'm shown there is no paretheses around the square root of x-5 is it still assumed to multiply?

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< sup >exponent< / sup > except, without the spaces.

So, let's break up (2 + (x - 4)^1/2)²
(2 + (x - 4)^1/2) * (2 + (x - 4)^1/2)

Also, I screwed up, since the coefficient should be 4, not 2.

FOIL:
First terms
(2 + (x - 4)^1/2) * (2 + (x - 4)^1/2)
Outer terms
(2 + (x - 4)^1/2) * (2 + (x - 4)^1/2)
Inner terms
(2 + (x - 4)^1/2) * (2 + (x - 4)^1/2)
Last terms
(2 + (x - 4)^1/2) * (2 + (x - 4)^1/2)

2*2 + 2*(x - 4)^1/2 + 2*(x - 4)^1/2 + (x - 4)
4 + 4(x - 4)^1/2 + x - 4
4(x - 4)^1/2 + x
3x - 8 = 4(x - 4)^1/2 + x
2x - 8 = 4(x - 4)^1/2
x/2 - 2 = (x - 4)^1/2

Square both sides:
(x/2 - 2)² = ((x - 4)^1/2)²
x²/4 - 2x + 4 = x - 4
x²/4 - 3x + 8 = 0

Multiply everything by 4 to get rid of the denominator:
x² - 12x + 32 = 0

Different answer. (x - 8)(x - 4) = 0; therefore, x = 8, 4.

%%
(2x - 1) = (x - 5) + 6(x - 5)^1/2 + 9?

Oh okay now I understand. You explain things a lot better than my math teacher who claims we should ask more questions in class but whenever I ask her a question she ignores me and I'm in the front.

I was wondering if you had a chance to look at the other problem?

Quote:

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Originally Posted by Depression Moon

the sum of three consecutive positive even intergers is equal to two more than seven times half the number. What are the three even positive intergers?

To me I'm imagining it looks like this.

a + b + c = 2 + 7 x 1/2n what's a,b, and c?

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I guess it's a bit late, but here it is.

So, the problem is telling us that we have three consecutive, positive, even integers. This is all important information, because it tells us that:
A) our integers are positive; negative answers will be incorrect
B) they are consecutive, which means that each variable can be stated in terms of the others
C) even, which means that they're even
D) the ambiguous "number" is the same on both sides

Three unknowns, more than three equations. Our equation that we get from the problem statement looks like:
a + b + c = (7/2)*x + 2
... solving for x, which is what your initial guess was.

But we know from our statements above, that each of the integers can be stated in terms of the others, which means that b = something*a. c = something*b; therefore, even c can be expressed in terms of a. We know that they're even, which means that each number is offset by 2 in sequence. If they were regular consecutive integers, we could just add +1, instead.

a (our first integer)
b = (a + 2)
c = (b + 2)

Substitute in what we know about b, into the equation for c, so that we can get rid of variables.
c = ((a + 2) + 2)

So, our integers now look like a, (a + 2), and (a + 4). Put those back into the problem statement's equation:
a + (a + 2) + (a + 4) = (7/2)x + 2
3a + 6 = (7/2)x + 2
3a + 4 = (7/2)x
6a + 8 = 7x

If x is a given, then you could solve the problem to get a number for a. Then add 2 to a to get b, and add 4 to get c. Otherwise, you have two unknowns, and only one equation: you can't solve the problem. However, we've been told that it's the same number on both sides; therefore, x = a.

6a + 8 = 7a
8 = (7a - 6a)
a = 8

6*8 + 8 = 7*8
48 + 8 = 56?
a = 8, b = 10, c = 12

QED

Quote:

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But we know from our statements above, that each of the integers can be stated in terms of the others, which means that b = something*a. c = something*b; therefore, even c can be expressed in terms of a. We know that they're even, which means that each number is offset by 2 in sequence. If they were regular consecutive integers, we could just add +1, instead.

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This part seemed a little confusing to me with the formula before it looks like it's handled differently, but I appreciate the help that second question gave me the worst headache ever.

What exactly is confusing you, then? How I got each variable to be expressed in terms of the variable a?