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Mazyrski [523]
3 years ago
8

Show that the roots of the equations

Mathematics
1 answer:
harkovskaia [24]3 years ago
3 0

Answer:

The roots are real and distinct.

Step-by-step explanation:

Given the following equation:

x^{2} + kx - k -2 =0

In this problem, a = 1, b = k and c = -k - 2

The discriminant is b² - 4ac, and for the roots to be real and distinct, it must be at least or greater than 0.

We get,

(k)²- 4(1)(-k - 2) = 1 - 4(-k - 2)

= k² + 4k + 8

Let's check:

At k = -2,

(-2)^{2} + 4(-2) + 8 = 4 - 8 + 8 = 4




At k = 0,

(0)^2 + 4(0) + 8 = 8

At k = -100,

(-100)^2 + 4(-100) + 8 = 10,000 - 400 + 8 = 9608

Therefore, we can conclude that for all values of k, the roots are real and distinct.

This has been a long way in answering this question, so it would be great if you could mark me as brainliest

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Solve the following quadratics. State the FACTORS AND SOLUTIONS. 1. 2x^2 - 7x + 3 2. 3x^2 + 7x +2
tekilochka [14]

Answer:

1. x = 3, 1/2 (solutions); (x - 3)(2x - 1) (factors)

2. x = -1/3, -2 (solutions); (3x + 1)(x + 2) (factors)

Step-by-step explanation:

<u>1. 2x^2 - 7x + 3</u>

To solve problem 1, you will need to identify your a, b, and c values in this quadratic function.

Since this problem is in standard form, it will be easy to identify these values. The standard form of a quadratic function is ax^2 + bx + c.

The a value is 2, the b value is -7, and the c value is 3 if we use our standard form and see which numbers are plugged into it.

Since we know that

  • a = 2
  • b = -7
  • c = 3

we can use the quadratic formula: x = \frac{-b~\pm~\sqrt{b^2~-~4ac} }{2a}

Substitute the a, b, and c values into the quadratic formula: x=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(3)} }{2(2)}

Now simplify using the laws of pemdas: x=\frac{7\pm\sqrt{(49)-(24)} }{4}

Simplify even further: x=\frac{7\pm\sqrt{(25)} }{4} \rightarrow x=\frac{7\pm (5) }{4}

Now split this equation into two equations to solve for x: x=\frac{12 }{4} ~~and~~ x=\frac{2 }{4}

12/4 can be simplified to 3, and 2/4 can be simplified to 1/2.

This means your solutions to problem 1 is 3, 1/2.

\boxed {x=3,\frac{1}{2} }

There is also another way to solve for the quadratic functions, and this was by factoring.

If you factor 2x^2 - 7x + 3 using the bottoms-up method, you will get (x - 3)(2x - 1).

After factoring, solving for the solutions is simple because all you have to do is set each factor to 0.

  • x - 3 = 0
  • 2x - 1 = 0

After solving for x by adding 3 to both sides, or by adding 1 to both sides then dividing by 2, you will end up with the same solutions: x = 3 and x = 1/2.

<u>2. 3x^2 + 7x + 2</u>

To save time I'll be using the bottoms-up factoring method, but remember to refer back to problem 1 (quadratic formula) if you prefer that method.

Factor this quadratic function using the bottoms-up method. After factoring you will have (3x + 1)(x + 2). These are your factors.

Now to solve for x and find the solutions of the quadratic function, you will set both factors equal to 0.

  • 3x + 1 = 0
  • x + 2 = 0

Solve.

<u>First factor:</u> 3x + 1 = 0

Subtract 1 from both sides.

3x = -1

Divide both sides by 3.

x = -1/3

<u>Second factor:</u> x + 2 = 0

Subtract 2 from both sides.

x = -2

Your solutions are x = -1/3 and x = -2.

\boxed {x = -\frac{1}{3} , -2}

7 0
3 years ago
Find the missing side of the triangle. A. 321−−−√ yd B. 221−−−√ yd C. 338−−√ yd D. 21−−√ yd
antoniya [11.8K]

Answer:

x = \sqrt{221} yd

Step-by-step explanation:

Use Pythagorean theorem to find x.

Thus, the sum of the square of the lengths of two legs of a right triangle equal the square of the hypotenuse, which is the longest side.

Thus,

x^2 = 11^2 + 10^2

x^2 = 121 + 100

x^2 = 221

x = \sqrt{221} yd

5 0
3 years ago
Ariel got 24 out of 30 questions correct on her
pentagon [3]
80%

24/30 = 24 divided by 30

When divided, you get 0.8

0.8 = 0.80

Therefore, the answer is 80%
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PLS ANSWER QUESTIONS 1,2,3
Dafna11 [192]

This is a mathematical word problem, so keep that in mind.

Hence, the reading for January 9 is -1°F (Option B)

The reading for Friday in the middle of the month is +1°F  (Option C); and

<h3>What is a math word problem?</h3>

A mathematical word problem is a mathematical question that is stated in the form of a sentence. See the answers below.

<h3>What are the answers as given in the question?</h3>

The answers are given in a tabular form to enable proper comprehension:

Date                              Low-Temperature reading

  • January  2  -5°F (three degrees lower than that of Jan. 23)
  • January  9  -1°F (two degrees lower than the low January 16)
  • January  16   1°F (6 degrees higher than the first Friday day)
  • January  23   -2°F
  • January  30      5°F (Seven degrees higher than the Friday before)

Learn more about word problems at:
brainly.com/question/13818690
#SPJ1

7 0
2 years ago
You roll a fair 666-sided die. What is \text{P(roll a 2})P(roll a 2)P, left parenthesis, r, o, l, l, space, a, space, 2, right p
liraira [26]

Answer:0.5

Step-by-step explanation:

​  

8 0
3 years ago
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