What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0? Use u substitution and the quadratic formula to solve. x = StartFraction negative 4 plus-or-minus StartRoot 5 EndRoot Over 2 EndFraction
x = StartFraction negative 7 plus-or-minus StartRoot 5 EndRoot Over 2 EndFraction
x = –7 and x = –2
x = –1 and x = 4

x = -7 ±√5  /2

That is;

x = StartFraction negative 7 plus-or-minus StartRoot 5 EndRoot Over 2 EndFraction

Step-by-step explanation:

To find the solutions to the equation, we will follow the steps below

(2x + 3)² + 8(2x + 3) + 11 = 0 (1)

let u = 2x + 3

we will replace 2x+ 3 by  u  in equation (1)

u² + 8u + 11 = 0  (2)

we will solve equation (2) using  the formula method

a=1   b=8     and c=11

u = -b ± √b² - 4ac  /2a

u = -8 ±√(-8)² - 4(1)(11)    /2(1)

u = -8 ±√64 - 44    /2

u = -8 ±√20    /2

u = -8/2  ± √20/2

u= -4 ± √20/2

u= -4 ± √5×4   /2

u= -4 ± 2√5 /2

u= -4 ± √5   (30

but u = 2x + 3

Substitute u =2x+3 in equation (3)

2x + 3 = -4 ± √5

subtract 3 from both-side of the equation

2x + 3-3 = -4 -3 ± √5

2x = -7±√5

Divide both-side of the equation by 2

2x/2 = -7 ±√5  /2

x = -7 ±√5  /2

the answer is b

Step-by-step explanation:

i took the test

, and

for

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote , , , so that we can call

note the following:

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from , to get to , we add

from , to get to , we add

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

, and for

the surface area is 124cm²

step-by-step explanation:

length: 4cm

width: 9cm

height:   2cm