Okay here are a few things i need help with please...
1. The roots of the equation is equal. Calculate the roots of the equation.
I keep getting that the roots of this equation doesnt exist...
2. and where is a constant and is a factor of . Calculate
I got
3. We have a graph and we are asked to give the equation of the graph symmetrical to the above one, with respect to the line .
I was thinking it could be: ?
But i'm really not sure.
4. The function is defined by
Determine if
I said that:
But i dont know what to do next.
Thanks for any help guys.
1. The roots of the equation x^2 -kx +k + 3 = 0 is equal. Calculate the roots of the equation.
You mean the roots are equal. Then the equation is a perfect square, in the form (x-a)^2.
So it can be done by "completing the square", or by the Qudratic Formula.
In the x = {-b +,-sqrt[b^2 -4ac]} / 2a,
if the [b^2 -4ac] is zero, then x = -b/(2a) only. Two of them (equal), because a quadratic always has two roots.
x^2 -kx +k + 3 = 0 --------------(i)
x^2 -kx +(k+3) = 0
The [b^2 -4ac] is k^2 -4(1)(k+3).
equate that to zero,
k^2 -4k -12 = 0
Factor that,
(k-6)(k+2) = 0
k = 6 or -2.
When k = 6, in (i),
x^2 -6x +6 +3 = 0
x^2 -6x +9 = 0
(x-3)^2 = 0
x = 3
When k = -2, in (i),
x^2 +2x -2 +3 = 0
x^2 +2x +1 = 0
(x+1)^2 = 0
x = -1
Therefore, for the two roots to be equal, the equation (i) has 2 roots of 3, or 2 roots of -1. ----answer.
2.
g(x) = 2x +1 and p(x) = g(x+k) where k is a constant and (x-1) is a factor of p(x). Calculate k.
p(x) = 2(x+k) +1 ------(i)
if (x-1) is a factor of p(x), then the other factor must have a degree of zero because p(x) is linear only (degree of 1) and (x-1) is of degree 1 already. So the other factor is a constant, say "a".
a(x-1) = 2(x+k) +1
ax -a = 2x +2k +1
ax -a = 2x +(2k +1)
For them to be equal, it must be that
ax = 2x ---------------(1)
-a = 2k +1 -----------(2)
ax = 2x
So, a = 2
Substitute that into (2),
-2 = 2k +1
-2 -1 = 2k
k = -3/2 ------------------------answer.
3. We have a graph y = -2x +6 and we are asked to give the equation of the graph symmetrical to the above one, with respect to the line y = x.
I was thinking it could be: x = -2y +6?
But i'm really not sure.
y = -2x +6
y = -2(x-3)
Swap x and y,
x = -2(y -3)
-x/2 = y -3
-x/2 +3 = y
or,
y = -(1/2)x +3 ----------------the equation of that line.
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4. The function f is defined by f(x) = x^(log[100x])
Determine x if f(x) = 1000.
I said that: 10^3 = x^(2x)
But log[100x] is not equal to 2x.
It is equal to log[100] +log[x] = 2 + log[x].
x^(log[100x]) = 1000
Take the logs of both sides,
(log[100x])*log[x] = log[10^3]
(log[100] +log[x])*log[x] = 3
(2 +log[x])*log[x] = 3
2log[x] +(log[x])^2 = 3
For simplicity, let y = log[x],
2y +y^2 = 3
y^2 +2y -3 = 0
(y+3)(y-1) = 0
y = -3 or 1
Or,
log[x] = -3 or 1
log[x] = -3
x = 10^(-3) = 1/1000 --------------answer.
log[x] = 1
x = 10^1 = 10 ----------------------answer also.
We are given the following function:
And we are told that point F with co-ordinates of (2 ; -9) is a local minimum.
---
1. Show that and
I set point F into and got:
When they say show then it means something completely different than prove doesnt it? Does it mean i can just go ahead and set and into and see if the equation is true?
Equating to zero and using the quadratic formula we get
We have two solutions, but since we know that in a cubic graph when has a coefficient the minimum is farther right than the maximum (hopefully someone with more experience can explain this better ). Hence, the x coordinate of the minimum is:
So you have two simultaneous equations:
Show assumes you don't know that p=-5 and q=-4 and you have to work them out from scratch.
I sort of had the idea... OK, i was wrong! Good one DivideBy0!
EDIT: Hey no, wait, this morning i started with it, except that my formulas were just the other way round.
Like the square was (L - x), etc.
But i stopped halfway through it because i had work to do...
But i did it the way you did...
Well tomorrow is D-Day for me...
Anyway, I have one more question.
For which values of will have only one root? (There has been proven that and )
I've thought about using Delta, and equating it to zero, but that only applies to quadratic equations. This question only counts 2 marks, so the solution should be quite simple.