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Simora [160]
3 years ago
11

What are the solutions of the quadratic equation (x + 3)2 = 49? A. O x = 2 and x = - 10 B.O.x = 4 and x = -10 C.O x=-2 and x = -

16 D.O x = 40 and x = -58​
Mathematics
1 answer:
tatyana61 [14]3 years ago
8 0

Answer:

B. x = 4 and x = -10

Step-by-step explanation:

First, expand the equation

(x + 3)² = 49

(x + 3)(x + 3) = 49

x² + 6x + 9 = 49

Subtract 49 from both sides

x² + 6x -40 = 0

Factor

(x + 10)(x - 4) = 0

Solve

x + 10 = 0

x = -10

x - 4 = 0

x = 4

So, the correct answer is B. x = 4 and x = -10

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ivolga24 [154]
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</span>x^2 + 2x + 4= 0.  "Completing the square" seems to be the easiest way to go here:

rewrite   x^2 + 2x + 4  as   x^2 + 2x + 1^2 - 1^2 = -4, or
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or                                            x = -1 plus or minus i*sqrt(3)

This problem, like any other quadratic equation, has two roots.  Note that the fourth possible answer constitutes one part of the two part solution found above.
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4vir4ik [10]
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8 0
3 years ago
Find the product. (-2x^2 )^3 ·3 x
jeka94

Assignment: \bold{Solve \ \left(-2x^2\right)^3\cdot \:3x}

<><><><><>

Answer: \boxed{\bold{-24x^7}}

<><><><><>

Explanation: \downarrow\downarrow\downarrow

<><><><><>

[ Step One ] Rewrite \bold{\left(-2x^2\right)^3}

\bold{-2^3\left(x^2\right)^3}

[ Step Two ] Rewrite Equation

\bold{-2^3\cdot \:3x^6x}

[ Step Three ] Apply Exponent Rule

Note: \bold{Exponent \ Rule \ \:a^b\cdot \:a^c=a^{b+c}: \ x^6x=\:x^{6+1}=\:x^7}

\bold{-2^3\cdot \:3x^7}

[ Step Four ] Refine

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\bold{\rightarrow Rhythm \ Bot \leftarrow}

4 0
3 years ago
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Nana76 [90]

Answer:

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