Need assistance with maths q's! (1 Viewer)

[x.Exhaust.x]

Exam tomorrow and this is meant to be revision. Stress stress stress...

3. For what values of p will the equation 2x^2-px+p-2=0 have:

c) Unequal roots which are rational
d) Unequal roots which are irrational

4. For the line 2x+3y-12=0, what is the angle the line makes with the x-axis?

Tan inverse 4/6 = 33 degrees 41 minutes.

The answer says 146 degrees 19 minutes. Why do you subtract by 180? Any reasoning?

5. If P is the point (-4,6) and Q is the point (2,-2), what is the equation of the line through Q perpendicular to the x-axis?

6. Find the equation of the line that:

b) Is the perpendivular bisector of the interval joining (-1,-4) and (-3,-2).

y-y1/x-x1=y2-y1/x2-x1
y+4/x+1=-2/2
y+4/x+1=-1
y+4=-x-1
x+y+5=0
y=-x-5
m1xm2=-1
Let m2=1
y+4/x+1=1
y+4=x+1
x-y-3=0

The answer says x-y-1=0...

b) Find the equation of a line which passes through the point of intersection of the lines 2y-3x-6=6x+9y-14, which also passes through the point (1,-3)

I used the formula: (a1x+b1y+c1)+k(a2x+b2y+c2); where k is a constant.

In the end, I get 3x+y=0....wrong?

STILL IN NEED OF HELP FOR THE ABOVE QUESTIONS

 

[tommykins]

回复: Need assistance with maths q's!

Exam tomorrow and this is meant to be revision. Stress stress stress...

1. Solve 9^x-6.3^x-27=0

(3^2)^x-6.3^x-27=0
(3^x)^2-6.3^x-27
Let m=3^x
m^2-6m-27
(m-9)(m+3)=0
m=9 or -3
3^x=9
x=2
3^x=-3 (not possible to find x-value)
Therefore x=2 only

Any other values?

Sounds right.

2. (x+3)^4-10(x+3)^2=-9
Let (x+3)^2=m
m^2-10m+9=0
(m-9)(m-1)=0
m=9, m=1
(x+3)^2=9, x+3=3, x=0
(x+3)^2=1, x+3=1, x= -2

I only seem to get two values, but in the answers, it has 4 values...I'm lost.

(x+3)^2=9, x+3 = +- sqrt 9 = +-3 -> x + 3 = +-3 , thus x = 0 or -6
(x+3)^2=1, x+3 = +- 1, x = 2 or -4

3. For what values of p will the equation 2x^2-px+p-2=0 have:

a) Unreal roots

p^2 - 4(p-2)(2) = p^2 - 8(p-2) = p^2 -8p +16 = (p-4)^2

to have unreal roots, discriminant must <0 but (p-4)^2 < 0 isn't true as any number ^2 is > 0.

b) Equal roots

From a) we have (p-4)^2 = 0, p = 4

c) Unequal roots which are rational
d) Unequal roots which are irrational

No idea since I'm confused with the above.. .

Likewise, haven't done discriminant in ages.

4. For the line 2x+3y-12=0, what is the angle the line makes with the x-axis?

Tan inverse 4/6 = 33 degrees 41 minutes.

The answer says 146 degrees 19 minutes. Why do you subtract by 180?

The angle is measured anti clockwise.

6. Find the equation of the line that:

a) Has an angle of inclination of 135 degrees to the x-axis and a y intercept of 2.

tan135=-tan45=-1

x/-1 + y/2 = 1
2x-y/-2=1
2x-y=-2
2x-y+2=0
y=2x+2
y=x+1

The answer says y=-x+2, did I follow an invalid procedure?

We know the m must be -1, so we use y = mx + b, m = -1 so y = -x + b, when x = 0 , y = 2, thus y = -x + 2

b) Is the perpendivular bisector of the interval joining (-1,-4) and (-3,-2).

Find gradient of the 2 points. Then find the midpoint of the 2 points.

m1m2 = -1, so solve for m2 (the bisector line)

y - midpointy = m2(x-midpointx) then just solve

6a) What is the angle between the two lines 2y-3x=6 and 6x+9y=14?

m1=3/2, m2= -2/3

Tan theta= |m1-m2/1+m1m2|
Tan theta= |3/2+2/3 / 1+ (3/2)(-2/3)|

The denominator is 0, therefore it's an infinite angle? lol.

m1m2 = -1 thus it is 90 degrees. reason it didn't work is you can't tan 90.



I did all the ones I could do in my head, sorry I couldn't answer it all.

 

[JSBboag]

6a) use the formula:
y-y(1) = m(x-x(1))

1.) you need to find the common y-values and x-values of the two lines - to give you a point (x,y) -?
2.) Then, find the intersection of the angle (it is not an infinite angle)
3.) sub into equation
b.)
1.) to the same thing but this time you are given the point (1,-3)
2.) also, don't forget to flip 'm' around and change the sign to satify the insection of the line.

 

[x.Exhaust.x]

Re: 回复: Need assistance with maths q's!

The angle is measured anti clockwise.

Thanks tommy and co. Tommy, why is the answer measured anti-clockwise? I drew a straight line, made the angle theta, and got my answer. Any special reason?

 

[tommykins]

回复: Re: 回复: Need assistance with maths q's!

Thanks tommy and co. Tommy, why is the answer measured anti-clockwise? I drew a straight line, made the angle theta, and got my answer. Any special reason?

It's weird though as the formula is meant to find acute angles.

But if there was a logical reasoning for me, I'd say you shoudl look at your ASTC graph, how you measure anti clockwise for the angle.

I'm not sure either, sorry.

 

[Aplus]

Re: 回复: Need assistance with maths q's!

Thanks tommy and co. Tommy, why is the answer measured anti-clockwise? I drew a straight line, made the angle theta, and got my answer. Any special reason?

ATSC diagram is measured anti clockwise, as aforementioned.

 

[lolokay]

Exam tomorrow and this is meant to be revision. Stress stress stress...

3. For what values of p will the equation 2x^2-px+p-2=0 have:

c) Unequal roots which are rational
d) Unequal roots which are irrational

the discriminant is p^2 - 8p + 16 = (p -4)^2
putting this into the quadratic formula you get (p +- sqrt(pi4)^2)/4
= (p +- (p - 4))/4
(2p - 4)/4 and 4/4 = 1, so 1 is always a root. (check by subbing in x=1)
so you can see that the other root will be rational if p is rational (unequal if p=/=4), irrational if p is irrantional (though you only have one irrational root)

4. For the line 2x+3y-12=0, what is the angle the line makes with the x-axis?

Tan inverse 4/6 = 33 degrees 41 minutes.

The answer says 146 degrees 19 minutes. Why do you subtract by 180? Any reasoning?

as has been mentioned, you generally measure angles anti-clockwise - however, I think 33'41' should still be right since it answers the question, not sure though. I would just specify that this is the acute angle made by the line and x axis.

5. If P is the point (-4,6) and Q is the point (2,-2), what is the equation of the line through Q perpendicular to the x-axis?

that's an odd question. Is it just y = -2?

6. Find the equation of the line that:

b) Is the perpendivular bisector of the interval joining (-1,-4) and (-3,-2).

y-y1/x-x1=y2-y1/x2-x1
y+4/x+1=-2/2
y+4/x+1=-1
y+4=-x-1
x+y+5=0
y=-x-5
m1xm2=-1
Let m2=1
y+4/x+1=1
y+4=x+1
x-y-3=0

The answer says x-y-1=0...

not too sure what you're working is, but..
the midpoint is (-2, -3)
the gradient of those points is -1, so perpendicular gradient is 1
y +3 = x +2
x - y - 1 = 0

b) Find the equation of a line which passes through the point of intersection of the lines 2y-3x-6=6x+9y-14, which also passes through the point (1,-3)

I used the formula: (a1x+b1y+c1)+k(a2x+b2y+c2); where k is a constant.

In the end, I get 3x+y=0....wrong?

STILL IN NEED OF HELP FOR THE ABOVE QUESTIONS

you got that right