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secondary 3 | A Maths
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Hi, does anyone know how to do question 5 and 6? I know that in q5 there aren’t any intersections, but I tried to solve it and I got a cubic equation and I just got stuck trying to find coordinates A. Thank you in advance
Date Posted: 4 years ago
Views: 252
We have to Guess a number in that polynomial
Basically for that equation I would expect a cubic equation in x since you are trying to solve an x^1 with an x^-2.
From there we just solve the polynomial as though it were a polynomial question from the start.
From there we just solve the polynomial as though it were a polynomial question from the start.
2x + 1 = 3/x²
2x³ + x² = 3
2x³ + x² - 3 = 0
sub values of x to try.
sub x = 1,
2(1)³ + (1)² - 3 = 2 + 1 - 3 = 0
By the factor theorem, (x - 1) is a factor.
(x - 1)(2x² + 3x + 3) = 0
You can either use long division or do it mentally
so x = 1 or 2x² + 3x + 3 = 0
For 2x² + 3x + 3 = 0,
Discriminant(b² - 4ac)
= 3² - 4(2)(3)
= 9 - 24
= -15 < 0
Since discriminant < 0, there are no real roots of x for 2x² + 3 = 0
So x = 1 is the only real x-coordinate of intersection.
Or
2x² + 3x + 3 = 0
Using quadratic formula,
x = ( -3 ± √(3² - 4(2)(3) ) / (2(2) )
x = (-3 ± √-15 )/4
There are no real roots of x here as √-15 gives a non-real/imaginary number.
Therefore,
y = 2(1) + 1 = 3
Or
y = 3/1² = 3
2x³ + x² = 3
2x³ + x² - 3 = 0
sub values of x to try.
sub x = 1,
2(1)³ + (1)² - 3 = 2 + 1 - 3 = 0
By the factor theorem, (x - 1) is a factor.
(x - 1)(2x² + 3x + 3) = 0
You can either use long division or do it mentally
so x = 1 or 2x² + 3x + 3 = 0
For 2x² + 3x + 3 = 0,
Discriminant(b² - 4ac)
= 3² - 4(2)(3)
= 9 - 24
= -15 < 0
Since discriminant < 0, there are no real roots of x for 2x² + 3 = 0
So x = 1 is the only real x-coordinate of intersection.
Or
2x² + 3x + 3 = 0
Using quadratic formula,
x = ( -3 ± √(3² - 4(2)(3) ) / (2(2) )
x = (-3 ± √-15 )/4
There are no real roots of x here as √-15 gives a non-real/imaginary number.
Therefore,
y = 2(1) + 1 = 3
Or
y = 3/1² = 3
For 6,
Rewrite the equation in the form
(x - a)² + (y - b)² = r² first.
Then (a,b) is your centre.
Any chord of the circle can be perpendicularly bisected by a certain radius of the circle.
Since the perpendicular bisector passes through the centre, use the points (a,b) and M(2,3) to find its gradient as they lie on it.
Then since it's perpendicular to the chord,
Gradient of chord = -1/gradient of perpendicular bisector of chord
So for y = mx + c, you already know m now.
Since M(2,3) lies on the chord as well ,
Sub the coordinates in to find c. Then you have your answer.
Rewrite the equation in the form
(x - a)² + (y - b)² = r² first.
Then (a,b) is your centre.
Any chord of the circle can be perpendicularly bisected by a certain radius of the circle.
Since the perpendicular bisector passes through the centre, use the points (a,b) and M(2,3) to find its gradient as they lie on it.
Then since it's perpendicular to the chord,
Gradient of chord = -1/gradient of perpendicular bisector of chord
So for y = mx + c, you already know m now.
Since M(2,3) lies on the chord as well ,
Sub the coordinates in to find c. Then you have your answer.
@Eric Nicholas K and @J thank you so much! I can visualise it now
Welcome
Hi Ell, please check the edited working above. Have edited some errors.
Additional info to understand why there are no other points of intersection :
The x and y axis are asymptotes of the curve y = 3/x² (x tends to 0 but doesn't reach 0 as y tends to infinity, vice versa) so the curve never intersects them.
And there is no part of the curve on and below the x-axis (i.e no values of y ≤ 0).
So the curve doesn't intersect the portion of y = 2x + 1 for those regions.
As for above the x-axis, the intersection is only at A for the region between the curve.
From point A, as x tends to infinity, y is strictly increasing and tends to infinity for y = 2x + 1.
But for y = 3/x², as x tends to infinity, y is strictly decreasing and tends to 0.
So the values of y are heading in diametrically opposite directions as thus there are no more intersections.
The x and y axis are asymptotes of the curve y = 3/x² (x tends to 0 but doesn't reach 0 as y tends to infinity, vice versa) so the curve never intersects them.
And there is no part of the curve on and below the x-axis (i.e no values of y ≤ 0).
So the curve doesn't intersect the portion of y = 2x + 1 for those regions.
As for above the x-axis, the intersection is only at A for the region between the curve.
From point A, as x tends to infinity, y is strictly increasing and tends to infinity for y = 2x + 1.
But for y = 3/x², as x tends to infinity, y is strictly decreasing and tends to 0.
So the values of y are heading in diametrically opposite directions as thus there are no more intersections.
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For qn (5). There are three intersections. For the first two, you need only sub into x=0 and y=0 as the line passes through both the x-axis and y-axis. The harder one to find is the intersection between the tangent and the curve. All you need to do is to do simultaneous equation. At the cubic part, I'm pretty sure using the calculator is fine. You can solve 2x^3 + x^2 - 3 =0 using mode/3/then 4. Not sure if it is the same for you as my calc is from a fewer versions back. You should get x=1 and two other that are nullified which is right since there is only one intersection between them. Sub in x=1 to either eqn and you'll get y=3. So A is (1,3)
Date Posted:
4 years ago
Finding the intersection points between y=2x+1 and the axes is not required as question asks for coordinates of A , i.e points of intersection between the two functions.
As for the calculator part, it can be used to find the root fast/check if root found by inspection is correct, but working must still be shown.
i.e sub x = 1 and find that f(1) = 0 so (x -1) is a factor by the Factor Theorem. Not doing so and the subsequent factorisation will result in loss of marks.
Since the question asks for any other points of intersection , it has to be shown that there are no other real roots other than x = 1 by showing discriminant < 0 for 2x² + 3 = 0. Then it can be concluded that there are no other intersections.
As for the calculator part, it can be used to find the root fast/check if root found by inspection is correct, but working must still be shown.
i.e sub x = 1 and find that f(1) = 0 so (x -1) is a factor by the Factor Theorem. Not doing so and the subsequent factorisation will result in loss of marks.
Since the question asks for any other points of intersection , it has to be shown that there are no other real roots other than x = 1 by showing discriminant < 0 for 2x² + 3 = 0. Then it can be concluded that there are no other intersections.
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