boredsatan
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- 1998
could you please explain how it would be done?No. The inverse should involve a square root.
could you please explain how it would be done?No. The inverse should involve a square root.
What was your progress on this problem?√(a-x) and a-x^2 intersect at x = 1
Find the possible values for a
set the equations equal to each other and then got stuckWhat was your progress on this problem?
sqrt(a-x) = a-x^2√(a-x) and a-x^2 intersect at x = 1
Find the possible values for a
a-x = a^2 - 2ax^2 + x^4sqrt(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
anyone?a-x = a^2 - 2ax^2 + x^4
how did you get this step?
Try and understand the problem and translate it into mathematics.How should I go about solving complex word problems?
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^2a-x = a^2 - 2ax^2 + x^4
how did you get this step?
but howcome it says x^4?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
Because it's a-x = (a-x^2)^2but howcome it says x^4?
I think so but the question is worded weirdly.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?
Noshow 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?
How would it be done? Kind of confused
Consider cases of a > 0 and a < 0.How would it be done? Kind of confused