This is the equation which you can add:
![[c + (3(c + 2))] + [5c]](https://tex.z-dn.net/?f=%5Bc%20%2B%20%283%28c%20%2B%202%29%29%5D%20%2B%20%5B5c%5D)
Simplify:
![[c + 3c + 6] + [5c] = 9c + 6](https://tex.z-dn.net/?f=%5Bc%20%2B%203c%20%2B%206%5D%20%2B%20%5B5c%5D%20%3D%209c%20%2B%206)
And, that's your answer!
Answer:
Step-by-step explanation:
13.
<h3>Given</h3>
<u>Quadratic equation</u>
- 4x² - 3x - 4 = 0
- With the roots of α and β
<h3>To Find </h3>
- The quadratic equation with roots of 1/(3α) and 1/(3β)
<h3>Solution</h3>
<u>The sum and the product of the roots of the given equation:</u>
- α + β = -b/a ⇒ α + β = -(-3)/4 = 3/4
- αβ = c/a ⇒ αβ = -4/4 = - 1
<u>New equation is:</u>
- (x - 1/(3α))(x - 1/(3β)) = 0
- x² - (1/(3α) + 1/(3αβ))x + 1/(3α3β) = 0
- x² - ((3α + 3 β)/(3α3β))x + 1/(3α3β) = 0
- x² - ((α + β)/(3αβ))x + 1/(9αβ) = 0
- x² - (3/4)/(3(-1))x + 1/(9(-1)) = 0
- x² + 1/4x - 1/9 = 0
- 36x² + 9x - 4 = 0
===================
14.
<h3>Given</h3>
<u>Quadratic equation</u>
- 3x² +2x + 7 = 0
- With the roots of α and β
<h3>To Find </h3>
- The quadratic equation with roots of α + 1/β and β + 1/α
<h3>Solution</h3>
<u>The sum and the product of the roots of the given equation:</u>
- α + β = -b/a ⇒ α + β = -2/3
- αβ = c/a ⇒ αβ = 7/3
<u>New equation is:</u>
- (x - (α + 1/β))(x - (β + 1/α)) = 0
- x² - (α + 1/β + β + 1/α)x + (α + 1/β) (β + 1/α) = 0
- x² - (α + β + (α + β)/αβ )x + αβ + 1/αβ + 2 = 0
- x² - (-2/3 - (2/3)/(7/3))x + 7/3 + 1/(7/3) + 2 = 0
- x² - (-2/3 - 2/7)x + 7/3 + 3/7 + 2 = 0
- x² + (14 + 6)/21x + (49 + 9 + 42/21) = 0
- x² + 20/21x + 100/21 = 0
- 21x² + 20x + 100 = 0
Answer:
x<u><</u> -7
Step-by-step explanation:
sorry I don't know the answer but i hope you get well soon
1/10 i believe... i hope it helps