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sleet_krkn [62]
3 years ago
11

Write the quadratic equation in general form and then choose the value of "b."

Mathematics
1 answer:
eduard3 years ago
3 0
The answer is b. 11. 

You can find this by multiplying the two parenthesis together. In order to do this, you can use a multiplying process known as FOIL (First, Outer, Inner, Last), where you multiply parts of the parenthesis in this order to make sure you do so properly. 

First Numbers:
2x * x = 2x^2

Outer Numbers: 
2x * 6 = 12x

Inner Numbers: 
-1 * x = -1x

Last Numbers: 
-1*6 = -6

Now place them all in that order and simplify. 
2x^2 + 12x - x - 6 ---> simplify x terms
2x^2 + 11x - 6

Since the middle term (the one with the x value) has a coefficient of 11, that gives us a b value of 11. 
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The roots are x = 2

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5 0
2 years ago
Find the smallest possible value of$$\frac{(y-x)^2}{(y-z)(z-x)} \frac{(z-y)^2}{(z-x)(x-y)} \frac{(x-z)^2}{(x-y)(y-z)},$$where $x
Harlamova29_29 [7]

Answer:

\frac{(y-x)^2}{(x-y)^2} \frac{(x-z)^2}{(z-x)^2}\frac{(z-y)^2}{(y-z)^2} =1

And on this case the samllest possible value would be 1

Step-by-step explanation:

For this case we have the following expression:

\frac{(y-x)^2}{(y-z)(z-x)} \frac{(z-y)^2}{(z-x)(x-y)} \frac{(x-z)^2}{(x-y)(y-z)}

And we can rewrite this expression like this:

\frac{(y-x)^2}{(x-y)^2} \frac{(x-z)^2}{(z-x)^2}\frac{(z-y)^2}{(y-z)^2}

For this case is important to remember the following property from algebra:

(a-b)^2 = a^2 -2ab + b^2

(b-a)^2 = b^2 - 2ab + a^2

On this case we can see that (a-b)^2 = (b-a)^2

So then (y-x)^2 = (x-y)^2 , (x-z)^2= (z-x)^2, (z-y)^2 =(y-z)^2

So then we can simplify all the expression and we got this:

\frac{(y-x)^2}{(x-y)^2} \frac{(x-z)^2}{(z-x)^2}\frac{(z-y)^2}{(y-z)^2} =1

And on this case the samllest possible value would be 1

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3 years ago
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