3 years ago

(x^2+4)^2−11(x^2+4)+24=0find one solution for this equation

Mathematics

1 answer:

fomenos3 years ago

4 0

Answer:

Step-by-step explanation:

(x²+4)²-11(x²+4)+24=0

put x²+4=y

y²-11y+24=0

y²-3y-8y+24=0

y(y-3)-8(y-3)=0

(y-3)(y-8)=0

y=3,8

so x²+4=3

x²=-1=i²

x=±i

or x²+4=8

x²=4

x=±2

You might be interested in

A package of 6 pairs of insulated socks costs $32.34. What is the unit price of the pairs of socks?

ANTONII [103]

32.34/6 = $5.39 per sock

5 0

3 years ago

BRAINLIEST TO BEST ANSWER :D

Temka [501]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

39. Which phrase is represented by 2(9-n)+3 ?

Leona [35]

Answer:

Step-by-step explanation:

It correctly describes it

7 0

3 years ago

Read 2 more answers

Chloe has enough sand to fill a sandbox with an area of 36 square units. She wants the outer edges of the sandbox to use as litt

sergiy2304 [10]

Answer:

The dimensions of the box is 6 × 6 units.

Step-by-step explanation:

Chloe fill a sand box whose area is 36 square units.

Let length be x units and width be y units.

Area of box = 36

$xy=36\\y=\dfrac{36}{x}\ \ \ ...(i)$

Perimeter of box, $P=2(x+y)$

$P=2(x+\dfrac{36}{x})\ \ \ \ \ \ \ \ [\text{ From }(i)]$

Differentiate w.r to x

$\dfrac{dP}{dx}=2(1-\dfrac{36}{x^2})$

For critical point, $\dfrac{dP}{dx}=0$

$\therefore 2(1-\dfrac{36}{x^2})=0$

$\Rightarrow x=\pm6$

x can't be negative.

$\dfrac{d^2P}{dx^2}=\dfrac{144}{x^3}\\\dfrac{d^2P}{dx^2}|_{x=6}=\dfrac{144}{216}>0(\text{min})$

$P_{min}=24$ units

Hence, Length = 6 units and width = 6 units

5 0

3 years ago

2+2=????????????????????????????????

arlik [135]

Answer:

4 ;P !!!!!!!!!!!!!!!!!!!!!!

8 0

3 years ago