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4 years ago

Solve x 2 - 4x - 7 = 0 by completing the square. What are the solutions?

Mathematics

1 answer:

Bingel [31]4 years ago

3 0

Answer:

x = 2±sqrt(11)

Step-by-step explanation:

x^2 -4x - 7 =0

Add 7 to each side

x^2 -4x -7+7 =0+7

x^2 -4x =7

To complete the square take the coefficient of the x term and divide by 2 and then square it. Add it to both sides

-4 /2 = -2 then square

(-2)^2 = 4

Add 4 to both sides

x^2 -4x+4 =7+4

x^2 -4x+4 = 11

Factor the left side into (x-b/2)^2

(x-2)^2 =11

Take the square root of each side

sqrt((x-2)^2) =±sqrt(11)

x-2 = ±sqrt(11)

Add to to each side

x-2+2 = 2±sqrt(11)

x = 2±sqrt(11)

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m = sqrt(a^2+b^2)
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Answer: 20

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3 years ago

Marissa bought x shirts that cost $19.99 each and y pairs of shorts that cost $14.99 each. The next day she went back to the sto

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Answer:

Step-by-step explanation:

Step one:

given data

day one

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4 more pairs of shorts cost $14.99 each.

Step two:

the total number of shirts is

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and the total number of shorts is

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19.99(x+3)+ 14.99(y+4)

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3 years ago

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Step-by-step explanation:

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3 years ago

Consider the graph of the cosine function shown below.

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Answer:

a. Amplitude: 4

Period: 1

b. The maximum values occur at $\theta=0,1,2$

The minimum values occur at $\theta=\frac{1}{2},\frac{3}{2}$

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Step-by-step explanation:

We can see from the graph that;

$-4\le f(x) \le 4$ .

This implies that;

$|A|=4$ .

The amplitude is 4.

The function also completed one full cycle on the interval [0,1]

The period is

$T=|1-0|=1$

The maximum values occur at $\theta=0,1,2$

The minimum values occur at $\theta=\frac{1}{2},\frac{3}{2}$

The zeros occur at $\theta=\frac{1}{4},\frac{3}{4},\frac{5}{4},\frac{7}{4}$

This is where the graph intersected the x-axis.

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3 years ago