For what value of m is the expression x4+mx2+64 a perfect square?
1 answer:
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The pattern is that the numbers in the right-most and left-most squares of the diamond add to the bottom square and multiply to reach the number in the top square.
For example, in the first given example, we see that the numbers 5 and 2 add to the number 7 in the bottom square and multiply to the number 10 in the top square.
Another example is how the numbers 2 and 3 in the left-most and right-most squares add up to the number 5 in the bottom square and multiply to the number 6 in the top square.
Using this information, we can solve the five problems on the bottom of the paper.
a) We are given the numbers 3 and 4 in the left-most and right-most squares. We must figure out what they add to and what they multiply to:
3 + 4 = 7
3 x 4 = 12
Using this, we can fill in the top square with the number 12 and the bottom square with the number 7.
b) We are given the numbers -2 and -3 in the left-most and right-most squares, which again means that we must figure out what the numbers add and multiply to.
(-2) + (-3) = -5
(-2) x (-3) = 6
Using this, we can fill the top square in with the number 6 and the bottom square with the number -5.
c) This time, we are given the numbers which we typically find by adding and multiplying. We will have to use trial and error to find the numbers in the left-most and right-most squares.
We know that 12 has the positive factors of (1, 12), (2,6), and (3,4). Using trial and error we can figure out that 3 and 4 are the numbers that go in the left-most and right-most squares.
d) This time, we are given the number we find by multiplying and a number in the right-most square. First, we can find the number in the left-most square, which we will call . We know that
, so we can find that
, or the number in the left-most square, is 8. Now we can find the bottom square, which is the sum of the two numbers in the left-most and right-most squares. This would be
. The number in the bottom square is
.
e) Similar to problem c, we are given the numbers in the top and bottom squares. We know that the positive factors of 8 are (1, 8) and (2, 4). However, none of these numbers add to -6, which means we must explore the negative factors of 8, which are (-1, -8), and (-2, -4). We can see that -2 and -4 add to -6. The numbers in the left-most and right-most squares are -2 and -4.
Answer:
Step-by-step explanation:
Given : universal set in this diagram is the set of integers from 1 to 15.
Solution :
The intersection of odd integer,multiples of 3 and Factors of 15 are 3,15
The intersection of odd integer and Factors of 15 are 1,5
The intersection of odd integer,multiples of 3 is 9
The remaining multiples of 3 are 6,12
The remaining odd integers are 7,11,13
Now the remaining integers are 2,4,8,10,14 and these integers must be placed in the boxes outside the circles Since they does not belong any intersection or odd integer or factor of 15 .
Refer the attached figure for the answer.
