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Verify that the roots of 5x²- 6x -2 = 0 are <img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%20%2B%20%5Csqrt%7B19%7D%20%7D%7B5%7D%2

Mice21 [21]

Answer:

Proof below.

Step-by-step explanation:

<u>Quadratic Formula</u>

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

<u>Given quadratic equation</u>:

$5x^2-6x-2=0$

<u>Define the variables</u>:

a = 5
b = -6
c = -2

<u>Substitute</u> the defined variables into the quadratic formula and <u>solve for x</u>:

$\implies x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(5)(-2)}}{2(5)}$

$\implies x=\dfrac{6 \pm \sqrt{36+40}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{76}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{4 \cdot 19}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{4}\sqrt{19}}{10}$

$\implies x=\dfrac{6 \pm2\sqrt{19}}{10}$

$\implies x=\dfrac{3 \pm \sqrt{19}}{5}$

Therefore, the exact solutions to the given <u>quadratic equation</u> are:

$x=\dfrac{3 + \sqrt{19}}{5} \:\textsf{ and }\:x=\dfrac{3 - \sqrt{19}}{5}$

Learn more about the quadratic formula here:

brainly.com/question/28105589

brainly.com/question/27953354

3 0

1 year ago

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The area of a rectangle is square units. Factor this expression. Given your answer in part A, describe what you can conclude abo

Brrunno [24]

Answer:

The factored area of the rectangle=2(2w-5)

Area of a rectangle=length×width

Step-by-step explanation:

Area of rectangle is 4w - 10 square units

We have to factor the expression

Area of rectangle = 4w - 10

common factor=2

Area of rectangle = 2(2w - 5)

Thus the given expression is factored

Area of a rectangle=length × width

Area of the rectangle=2×(2w-5)

Thus, it is safe to say

The length=2 or (2w-5)

The width=(2w-5) or 2

3 0

3 years ago

Static and Reasoning:

Serggg [28]

Based on the data recorded by Isabella, it can be concluded the cube is rather fair.

<h3>How many times did Isabella get each number?</h3>

Based on the data, here are the results:

Getting a 1: 6 times
Getting a 2: 5 times
Getting a 3: 7 times
Getting a 4: 5 times
Getting a 5: 6 times
Getting a 6: 7 times

This implies, in total Isabella got the same number between five and seven times. For example, the number 2 was obtained 5 times, but the number 3 was obtained 7 times.

<h3>What can be concluded based on the results?</h3>

Even though Isabella did not get the same number of times each number, the dice is rather fair because by rolling the dice thirty six times you will obtain the same number at least five times.

Moreover, there is not a big difference in the number of times you obtain each number.

Learn more about dice in: brainly.com/question/23637540

#SPJ1

5 0

1 year ago

Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k

masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0

3 years ago

How can I solve (-12 =2 + k)

ZanzabumX [31]

You would want to go by reverse order of operations. This is a one step equation. You subtract 2 on both sides, in which you will get: -14 = k.

3 0

3 years ago

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