00:05
Okay, for number one, we have a rectangular garden.
00:07
So i'm going to sketch here, a rectangle.
00:10
And the length is four meters longer than the width.
00:14
So i'm going to represent my width as w.
00:17
So this will be w times 4.
00:19
The area is 60.
00:22
So area of a rectangle is length times width.
00:25
So we're going to have w times w plus 4 equals 60.
00:33
I'm going to distribute my w and get w.
00:35
Squared plus 4w equals 60.
00:39
I have a quadratic.
00:41
So i'm going to subtract 60 from both sides, giving me w squared plus 4w minus 60 equals 0.
00:51
So now to factor, i need to find two factors of 60, of negative 60 that give me a difference of positive.
01:00
So we know that 10 times 6 is 60, and 10 minus 6 is 4.
01:06
So i'm going to have here, w6, and then w and 10.
01:13
Now, and this equals 0.
01:16
So since we have a negative 60 here, one has to be positive and one has to be negative, we know that the difference is going to be a positive 4, so we want our larger number 10 to be plus.
01:28
So i'm going to have w minus 6 equals 0, w plus 10 equals 0.
01:37
If w minus 6 equals 0, then our width is 6.
01:43
If w plus 10 equals 0, then w is negative 10.
01:48
We cannot have a length of a side of a rectangle be negative, so that answer is eliminated.
01:54
So our width is 6.
01:56
If our length is 4 more than the width, then the length is going to be 6 plus 4, which is 10.
02:03
10.
02:05
Next up for number two, i'm going to solve for three consecutive integers such that the square of the third is 51 more than the sum of the first and the second.
02:15
So i'm going to label my integers as a.
02:19
So my first integer will be a.
02:22
Since they are consecutive, my second integer will be a plus 1.
02:28
And my third consecutive integer can be represented as a plus 2.
02:34
So the square of a the third, here is my third right here.
02:38
So i'm going to take a plus 2, that's an a, a plus 2, and square it.
02:46
So the square of the third is going to equal 51 more, so 51 plus the sum of the first and the second.
02:56
So this first is a, the second is a plus 1.
03:02
And that's an a, not a 9.
03:05
So i'm going to expand and this binomial, it's going to be a perfect square trinomial, so it will be a squared plus 4a plus 4.
03:19
Because if you have a binomial squared, when it expands, it follows the pattern of a squared plus 2, ab plus b squared.
03:29
So on the right side of the equation, i'm going to combine light terms.
03:32
51 plus 1 is 52.
03:35
A plus a is 2a.
03:46
Now we have a quadratic, so i want to get all of my values onto one side of the equal sign and set it equal to 0.
03:53
So i'm going to subtract 2a from both sides.
03:56
So i'm going to a squared plus 2a.
04:02
So i'm going to subtract 52 from both sides for minus 52 is going to be negative 48.
04:12
And that is going to equal 0.
04:15
Now i need to take i need to look and factor this.
04:19
I need to find two factors of 48 of negative 48 that have the sum of 2.
04:25
So the two factors i'm going to use are 6 and 8.
04:41
Since 48 is negative, i know that one of these has to be positive, one has to be negative.
04:48
Since my 2 in the middle here is positive, i know my larger number has to have the plus in front of it.
04:56
So i'm going to have have a minus 6 times a plus 8 equals 0.
05:07
If i set each of these equal to 0, i get a equals 6, and a equals negative 8.
05:20
The question says that we're dealing with three consecutive positive integers, so negative 8 is not a possible solution.
05:28
So my positive integers, starting with a, will be 6, 7, and eight.
05:36
And if you wanted to check your work, you can plug these back in and confirm that it all evens out.
05:43
And last but not least, let's do number three...