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Afina-wow [57]
3 years ago
10

Solve the quadratic equation for x. What is the product when the roots are multiplied? (x + 2)2 = 1

Mathematics
2 answers:
tekilochka [14]3 years ago
4 0

Regroup terms

2(x + 2) = 1

Divided both sides by 2

x + 2 = 1/2

Subtract 2 from both sides

x = 1/2 - 2

Simplify 1/2 - 2 to -3/2

<u>x = -3/2</u>

sveticcg [70]3 years ago
4 0

Answer:

x = -2 or 0; Product of roots = 0.  

Step-by-step explanation:

Product of roots:

        (x + 1)² = 1

x² + 2x + 1    = 1

x²  +2x         = 0

The standard form of a quadratic is a² + bx + c = 0.

In your equation, a = 1; b = 2; c = 0

Even without solving the equation, we know that the product of the roots is zero, because

Product of roots = c/a =0/1 = 0

Solution:

Factor the quadratic                x(x +2) = 0

Use zero product property     x = 0     x + 2 =  0

                                                              x       = -2

The roots are x = 0 and x = -2.

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Which expressions are equivalent to the expression shown below? Select all that apply. 6x + 3y – 2x + 4y – 2y + x + 5x A 10x + 5
yuradex [85]

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2 years ago
The function y=60/x relates y, the time in hours that it takes micheal to travel to his grandparents house,to x , his average sp
Andrews [41]

Answer:

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9.99x10 ^-2 + 1.11 × 10^-2 i dont have the same scientific calculator as the one from school. Would like to verify my math is co
Rufina [12.5K]
The answer should not depend on which machine or which pencil you use to
find it.  If you work a problem two different ways and get two different answers,
then at least one of them is wrong, and there's a pretty good chance that both
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A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be
irga5000 [103]

Answer:

The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

Step-by-step explanation:

Let C = (h,k) the coordinates of the center of the circle, which must be transformed into C'=(h', k') by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

(x-h)^{2}+(y-k)^{2} = r^{2}

Where r is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) <em>Center of the circle is translated 4 units to the right</em> (+x direction):

C''(x,y) = C(x, y) + U(x,y) (Eq. 1)

Where U(x,y) is the translation vector, dimensionless.

If we know that C(x, y) = (h,k) and U(x,y) = (4, 0), then:

C''(x,y) = (h,k)+(4,0)

C''(x,y) =(h+4,k)

2) <em>Reflection over the x-axis</em>:

C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)] (Eq. 2)

Where O(x,y) is the reflection point, dimensionless.

If we know that O(x,y) = (h+4,0) and C''(x,y) =(h+4,k), the new point is:

C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]

C'(x,y) = (h+4, 0)-(0,k)

C'(x,y) = (h+4, -k)

And thus, h' = h+4 and k' = -k. The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

<em />

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