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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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Jada solved the equation Negative StartFraction 4 over 9 EndFraction = StartFraction x over 108 EndFraction for x using the step
GalinKa [24]

Answer:

Jada should have multiplied both sides of the equation by 108.

Step-by-step explanation:

The question is incomplete. Find the complete question in the comment section.

Given the equation -4/9 = x/108, in order to determine Jada's error, we need to solve in our own way as shown:

Step 1: Multiply both sides of the equation by -9/4 as shown:

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It can also be solved like this:

Given -4/9 = x/108

Multiply both sides by 108 to have:

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-4/9 * 108 = 108x/108

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x = -48

Jada should have simply follow the second calculation by multiplying both sides of the equation by 108 as shown.

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How do you subtract when there are zero's at the end? For ex. 1000-325 how'd you borrow from 1000?
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6 0
3 years ago
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