Mathematics Marathon (1 Viewer)

[bored of sc]

Answer the question then post your own.

1) Solve for x by completing the square:

8x² - 5x - 104 = 0

 

[lyounamu]

Answer the question then post your own.

1) Solve for x by completing the square:

8x² - 5x - 104 = 0

x = 5+_SR(5^2 - 4 . 8 . -104)/16
= 3.93156842... OR -3.30656842...

Damn read the question wrongly. Let me try again.

8(x^2-5/8x-13) = 0
8(x^2 - 5/8x + 25/256 - 13 25/256)
8(x^2-5/8x + 25/256) -3353/32 = 0
8(x-5/16)^2 - 3353/32 = 0
8(x-5/16)^2 = 3353/32
(x-5/16)^2 = 3353/256
x-5/16 = +_3.619068...
x = 3.9315... or -3.30654...

By the way, I thought there was another question initially. When I came back, it was gone...

 

[clintmyster]

8x2 - 5x - 104 = 0
8(x^2 - 5/8x) = 104
8(x^2 - 5/8x + [-5/16]^2)=104 +(5/16)^2
8(x - 5/16)^2 = 104 x 25/256

 

[bored of sc]

By the way, I thought there was another question initially. When I came back, it was gone...

Yeah, it didn't work (no solutions).
P.S Type up a question!

 

[Norma.Jean]

8x^2-5x-104=0
8x^2 -5x =104
x^2 -5/8x + (5/16)^2 = 13+25/256
(x-5/16)^2= 13 25/256
x-5/16= +- (square root) 13 25/256
x= +- (square root) 13 25/256 + 5/16
x= 3.93(to 2 d.p.) or x=-3.30(to 2 d.p)

man that is such a horrible thing to write
sorry if twenty other people have posted up the answer during this time

 

[clintmyster]

daym i forgot to solve for x haha

 

[lyounamu]

Yeah, it didn't work (no solutions).
P.S Type up a question!

Hey where is the quesiton that you posted up in the other thread. I solved it but can you give me that question again so that I can type it here.

 

[bored of sc]

Differentiate y = 5x² - 3 using first principles.

 

[ratcher0071]

Differentiate y = 5x² - 3 using first principles.

y' = lim h->0 f(x+h)-f(x) / h
y' = lim h->0 5(x+h)²-3 - (5x²-3) / h
y' = lim h->0 5x² + 10hx +10h² -3 - 5x² + 3 /h
y' = lim h->0 10hx +10h² /h
y' = lim h->0 h(10x+10h) /h
y' = lim h->0 10x+10h
y' = 10x

Sorry Namu

 

[bored of sc]

y' = lim h->0 f(x+h)-f(x) / h
y' = lim h->0 5(x+h)²-3 - (5x²-3) / h
y' = lim h->0 5x² + 10hx +10h² -3 - 5x² + 3 /h
y' = lim h->0 10hx +10h² /h
y' = lim h->0 h(10x+10h) /h
y' = lim h->0 10x+10h
y' = 10x

Sorry Namu

Can you post up another question to keep the thread going?

 

[ratcher0071]

Can you post up another question to keep the thread going?

Find the x-coordinates of the points P and Q on y=(x-h)² + k such that the tangents at P and Q have gradients m and -m respectively?

is that too hard?

 

[tommykins]

y = x² - 2xh + h² + k
dy/dx = 2x - 2h
Let 2x - 2h = m
2(x-h) = m
x-h = m/2
x = m/2 + h

Now let 2x - 2h = -m
x-h = -m/2
x = -m/2 + h

 

[ratcher0071]

y = x² - 2xh + h² + k
dy/dx = 2x - 2h
Let 2x - 2h = m
2(x-h) = m
x-h = m/2
x = m/2 + h

Now let 2x - 2h = -m
x-h = -m/2
x = -m/2 + h

Good Work Tommy, seems right

 

[ratcher0071]

Find f'(x) and f''(x) of f(x)= x⁴-1 / x² + x -1

Hint:
let u=x⁴ - 1
let v=x² + x -1

 

[omniscience]

Don't know what f''(x) means but,

f(x)= x<SUP>4</SUP>-1 / x<SUP>2</SUP> + x -1
f'(x) = [(x² + x - 1)(4x³) - x⁴(2x + 1)] / (x² + x - 1)²
= (4x⁵ + 4x⁴ - 4x³) - (2x⁵ + x⁴) / (x² + x - 1)²
= (2x⁵ + 3x⁴ - 4x³) / (x² + x - 1)²

f''(x) means second derivative. Just differentiate it again.

 

[bored of sc]

Find f'(x) and f''(x) of f(x)= x⁴-1 / x² + x -1

Hint:
let u=x⁴ - 1
let v=x² + x -1

f(x)= x<SUP>4</SUP>-1 / x<SUP>2</SUP> + x -1

f'(x) = [(x² + x - 1).4x³ - (x⁴ - 1)(2x + 1)] / (x² + x - 1)²

*Expands and simplifies:
= (2x⁵ + 3x⁴ - 4x³ + 2x + 1) / (x⁴ + 2x³ - x² - 2x + 1)

 

[bored of sc]

f''(x) means second derivative. Just differentiate it again.

The quotient rule is:

dy/dx = (vu' - uv') / v²

right?

 

[x.Exhaust.x]

The quotient rule is:

dy/dx = (vu' - uv') / v²

right?

Yep.

 

[bored of sc]

f''(x) = [(x⁴ + 2x³ - x² - 2x + 1)(10x⁴ + 12x³ - 12x² + 2) - (2x⁵ + 3x⁴ - 4x³ + 2x + 1)(4x³ + 6x² - 2x - 2)] / (x⁴ + 2x³ - x² - 2x + 1)²

Someone very careful can expand and simplify.

 

[omniscience]

The quotient rule is:

dy/dx = (vu' - uv') / v²

right?

Yes. Your working out is correct from my point of view.