Help please (1 Viewer)

[crazykat3000]

How do you solve these types of questions because I can never get them:

Express 9x^2+4x+16 in the form (Ax+1)^2+B(x-2)+C

Then you have to find out what A, B and C are.

 

[Azreil]

(3x+1)^2 -2x + 15
=(3x+1)^2 - 2(x-2) + 11
A= 3 B=-2 C=11

 

[crazykat3000]

...??

 

[henry08]

I don't get it either.

 

[minijumbuk]

Expand out the right bit.

(Ax+1)^2+B(x-2)+C = A²x²+2Ax + 1 + Bx - 2B + C
= (A²)x² + (2A + B)x + (1 - 2B + C) (collect like terms)

Equate coefficients of terms x², x and the constants with the equation, ie. 9x² + 4x + 16
So A²= 9, (2A + B) = 4, (1 - 2B + C) = 16
________________
A = 3

2(3) + B = 4
B = -2

1 - 2(-2) + C = 16
C = 11

OR
________________
A = - 3

2(-3) + B = 4
B = 10

1 - 2(10) + C = 16
C = 35
________________

So,
(3x+1)² -2(x-2) + 11
or
(-3x+1)² + 10(x-2) + 35

 

[henry08]

I still don't get it, will go read up the textbook.

 

[hoochiscrazy]

Anyone that can help thx.

Find the area bounded by the curve y=sin 2x the x-axis and the lines x=0 and x=pie

please post working

 

[dolbinau]

Anyone that can help thx.

Find the area bounded by the curve y=sin 2x the x-axis and the lines x=0 and x=pie

please post working

Is the answer two?

 

[stormz89]

Anyone that can help thx.

Find the area bounded by the curve y=sin 2x the x-axis and the lines x=0 and x=pie

please post working

[-1/2cos 2x]

-1/2cos 2x0 = -1/2

-1/2cos 2pie = -1/2

which = 0

That ain't right. any help?

EDIT: Actually according to my textbook it is right because it's above the graph until pie/2 and is symmetrical as he falsl below the x axis so the area covered is zero.

 

[Charlesr]

I did it a different way, I've never seen a question like that.
I figured the (Ax + 1)**2 was the only way you would get a coefficent to x**2.
Therefore A = 3
Expand that bit to get 9x**2 + 6x +1
Then subtract that from the equation your aiming for
You get -2x + 15
SO u get: B(x-2) + C = -2x + 15
fiddle for a second and get B = -2 and C = 11
Might be confusing or might help who knows

 

[Graceofgod]

[-1/2cos 2x]

-1/2cos 2x0 = -1/2

-1/2cos 2pie = -1/2

which = 0

That ain't right. any help?

EDIT: Actually according to my textbook it is right because it's above the graph until pie/2 and is symmetrical as he falsl below the x axis so the area covered is zero.

I can't be bothered doing it right now. But sin2x crosses the axis at pi/2.
So you need to add the absolute values of the two parts, above and below axis to get the total area.

 

[hoochiscrazy]

Is the answer two?

yeah answers 2 can u post working?

 

[stormz89]

I can't be bothered doing it right now. But sin2x crosses the axis at pi/2.
So you need to add the absolute values of the two parts, above and below axis to get the total area.

which would equal one if you look at it like that.

I was looking at it as a particle of movement which doesn't include absolute value.

EDIT: how is the answer 2? my working says 1


remember the integration of sin ax is -1/a cos ax.

 

[zthoms01]

Hi,

As said before the graph cuts the axis at pi/2 so take the absolute value of the areas and add them together. As Sin 2x is the same above and below the axis just multiply the integral from 0 - pi/2 by 2.

So.....

y = sin 2x
x = 0
x = pi

Therefore:

Int(0 to pi) sin 2x dx = 2 Int(0 to pi/2) sin 2x dx = 2[-1/2 cos2x][0 to pi/2]

= 2[1/2 + 1/2]

= 2 * 1

= 2 As Reqd.

Hope this helps!

Reply if you need more help! :wave:

 

[stormz89]

Hi,

As said before the graph cuts the axis at pi/2 so take the absolute value of the areas and add them together. As Sin 2x is the same above and below the axis just multiply the integral from 0 - pi/2 by 2.

So.....

y = sin 2x
x = 0
x = pi

Therefore:

Int(0 to pi) sin 2x dx = 2 Int(0 to pi/2) sin 2x dx = 2[-1/2 cos2x][0 to pi/2]

= 2[1/2 + 1/2]

= 2 * 1

= 2 As Reqd.

Hope this helps!

Reply if you need more help! :wave:

I'm trying to understand, how did you get 2[1/2 + 1/2]? I understand the need to double it but why 1/2 + 1/2? -1/2cos2x where x = pi/2 is just 1/2 so where does the other half come in?

 

[hoochiscrazy]

Hi,

As said before the graph cuts the axis at pi/2 so take the absolute value of the areas and add them together. As Sin 2x is the same above and below the axis just multiply the integral from 0 - pi/2 by 2.

So.....

y = sin 2x
x = 0
x = pi

Therefore:

Int(0 to pi) sin 2x dx = 2 Int(0 to pi/2) sin 2x dx = 2[-1/2 cos2x][0 to pi/2]

= 2[1/2 + 1/2]

= 2 * 1

= 2 As Reqd.

Hope this helps!

Reply if you need more help! :wave:

Thanks got it now. Havnt done maths in like 2 weeks so all the info has fallen out of my head lol

 

[zthoms01]

Hi,

Sorry i skipped a couple of steps.

In the primitive 2[-1/2 cos2x][0 to pi/2]

That becomes 2[(-1/2) (cos(pi)) - ((-1/2)cos(0))] <--- The definition of a definite integral

= 2 [ (-1/2)(-1) - (-1/2)(1) ]
= 2 [ 1/2 - (-1/2) ]
= 2 [ 1/2 + 1/2 ]
= 2

 

[zthoms01]

No Problem =]

 

[stormz89]

oh ok I get it now,

 

[hoochiscrazy]

k i got another for you guys now lol

find dy/dx given that y=ln(2x+1/3x-7)