VCE Maths questions help (1 Viewer)

[boredsatan]

No. The inverse should involve a square root.

could you please explain how it would be done?

 

[boredsatan]

√(a-x) and a-x^2 intersect at x = 1
Find the possible values for a

 

[InteGrand]

√(a-x) and a-x^2 intersect at x = 1
Find the possible values for a

What was your progress on this problem?

 

[boredsatan]

What was your progress on this problem?

set the equations equal to each other and then got stuck

 

[pikachu975]

√(a-x) and a-x^2 intersect at x = 1
Find the possible values for a

sqrt(a-x) = a-x^2
a-x = a^2 - 2ax^2 + x^4
Sub x = 1
a - 1 = a^2 - 2a + 1
a^2 - 3a + 2 = 0
(a-2)(a-1) = 0
a = 1, 2

 

[boredsatan]

sqrt(a-x) = a-x^2
a-x = a^2 - 2ax^2 + x^4
Sub x = 1
a - 1 = a^2 - 2a + 1
a^2 - 3a + 2 = 0
(a-2)(a-1) = 0
a = 1, 2

a-x = a^2 - 2ax^2 + x^4
how did you get this step?

 

[boredsatan]

a-x = a^2 - 2ax^2 + x^4
how did you get this step?

anyone?

 

[boredsatan]

How should I go about solving complex word problems?

 

[InteGrand]

How should I go about solving complex word problems?

Try and understand the problem and translate it into mathematics.

 

[jathu123]

a-x = a^2 - 2ax^2 + x^4
how did you get this step?

By squaring both sides. By squaring the LHS, we get rid of the square root. When we square the RHS, we expand it (A+B)^2 = A^2+2AB+B^2

 

[boredsatan]

By squaring both sides. By squaring the LHS, we get rid of the square root. When we square the RHS, we expand it (A+B)^2 = A^2+2AB+B^2

but howcome it says x^4?

 

[boredofstudiesuser1]

but howcome it says x^4?

Because it's a-x = (a-x^2)^2

And so the B^2 part is (x^2)^2 which = x^4

 

[boredsatan]

A ball flies through the air following the equation y = 2 + 0.12x - 0.002^2. The ground to where the ball lands is modelled by y = 0.15x. The coordinates on the ground where the cannon ball lands is.
Do I need to find the point of intersection of these lines?

 

[pikachu975]

A ball flies through the air following the equation y = 2 + 0.12x - 0.002^2. The ground to where the ball lands is modelled by y = 0.15x. The coordinates on the ground where the cannon ball lands is.
Do I need to find the point of intersection of these lines?

I think so but the question is worded weirdly.

Basically if you're stuck with a question, try any method you can think of! Pretend it's like an exam and you have to try something, don't just give up easily and ask others.

Also if there's worked solutions DEFINITELY look at them to see how to do it. Then close the answers and retry the question.

 

[boredsatan]

The equation of a straight line which makes an angle of 30 degrees with the positive x axis and cuts the y axis at 5
is this working our right
y - y1=m(x-x1)
y-5=m(x-0)
m = tan theta
m = tan 30 degrees
m = 0.577
y - 5 = 0.577(x-0)
y = 0.577x + 5

 

[boredsatan]

show that the graph of f(x) = px^2+ q has no x intercept if pq>0
so ax^2 +c
a*c>0, therefore the y intercept will be higher than 0, therefore there is no x intercept
Is this above working out correct?

 

[InteGrand]

show that the graph of f(x) = px^2+ q has no x intercept if pq>0
so ax^2 +c
a*c>0, therefore the y intercept will be higher than 0, therefore there is no x intercept
Is this above working out correct?

No

 

[boredsatan]

No

How would it be done? Kind of confused

 

[InteGrand]

How would it be done? Kind of confused

Consider cases of a > 0 and a < 0.

 

[boredsatan]

show that y = 4-6x+2x^2+3x^3 has exactly one x intercept
factorise it
(x+2)(3x^2-4x+2)
b^2 - 4ac = 0
16-4(3)(2) =
16-24 = -8
Not sure how to finish it