Answer:
Step-by-step explanation:
Answer:
According to steps 2 and 4. The second-order polynomial must be added by 
and 
to create a perfect square trinomial.
Step-by-step explanation:
Let consider a second-order polynomial of the form 
, 
. The procedure is presented below:
1) (Given)
2) 
(Compatibility with addition/Existence of additive inverse/Modulative property)
3) 
(Compatibility with multiplication)
4) 
(Compatibility with addition/Existence of additive inverse/Modulative property)
5) 
(Perfect square trinomial)
According to steps 2 and 4. The second-order polynomial must be added by and to create a perfect square trinomial.
Answer:
The correct answer is D) (-2, -1)
Step-by-step explanation:
In order to solve this system of equations, start by multiplying the entire first equation by 2. Then add the two equations together. This will get the y's to cancel and allow you to solve for x.
-4x + 2y = -10
3x - 2y = 12
---------------------
-x = 2
x = -2
Now that we have the value for x, we can find y by plugging the x value into either equation.
-2x + y = -5
-2(2) + y = -5
-4 + y = -5
y = -1
Answer:
- 5(x +1.5)^2
- 10(x +1)^2
- 1/4(x +2)^2
- 3(x +5/6)^2
Step-by-step explanation:
When your desired form is expanded, it becomes ...
a(x +b)^2 = a(x^2 +2bx +b^2) = ax^2 +2abx +ab^2
This tells you the overall factor (a) is the leading coefficient of the given trinomial. Factoring that out, you can find b as the root of the remaining constant.
a) 5x^2 +15x +11.25 = 5(x^2 +3x +2.25) = 5(x +1.5)^2
b) 10x^2 +20x +10 = 10(x^2 +2x +1) = 10(x +1)^2
c) 1/4x^2 +x +1 = 1/4(x^2 +4x +4) = 1/4(x +2)^2
d) 3x^2 +5x +25/12 = 3(x^2 +5/3x +25/36) = 3(x +5/6)^2
_____
<em>Additional comment</em>
If you know beforehand that the expressions can be factored this way, finding the two constants (a, b) is an almost trivial exercise. It gets trickier when you're trying to write a general expression in vertex form (this form with an added constant). For that, you must develop the value of b from the coefficient of the linear term inside parentheses.
