X^2-4x+4
(x-2)(x-2) or (x-2)^2
Check:
-2-2=-4
-2*-2=4
Given AP is 21 ,18,15,...
First term = 21
Common difference = 18-21 = -3
Let an = -81
We know that
an = a+(n-1)d
⇛21+(n-1)(-3) = -81
⇛ 21-3n+3 = -81
⇛24-3n = -81
⇛ 24+81 = 3n
⇛ 105= 3n
⇛ n = 105/3
⇛ n = 35
35th term of the AP is -81.
<u>also</u><u> read</u><u> similar</u><u> questions</u><u>:</u> Which term of a AP 5 , 13 , 21 ,... is 181?
brainly.com/question/2406241?referrer
Which term of the AP:3, 8, 13, 18,...,is 78?
brainly.com/question/15380012?referrer
Answer:
The first set of consecutive even integers equals (8 , 6)
The second set is ( - 8 and - 6) which also works.
Step-by-step explanation:
Equation
(x)^2 + (x + 2)^2 = (x)(x + 2) + 52 Remove the brackets on both sides
Solution
x^2 + x^2 + 4x + 4 = x^2 + 2x + 52 Collect the like terms on the left
2x^2+ 4x+ 4 = x^2 + 2x + 52 Subtract right side from left
2x^2 - x^2 + 4x - 2x + 4 - 52 = 0 Collect the like terms
x^2 + 2x - 48 = 0 Factor
(x + 8)(x - 6) = 0
Answer
Try the one you know works.
x - 6 = 0
x = 6
Therefore the two integers are 6 and 8
6^2 + 8^2 = 100
6*8 + 52 = 100
So 6 and 8 is one set of consecutive even numbers that works.
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What about the other set.
x + 8 = 0
x = - 8
x and x + 2
- 8 and -8 + 2 = - 8, - 6
(- 8 )^2 + (- 6)^2 = 100
(-8)(-6) + 52 = 100
Both sets of consecutive numbers work.
