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To find points of intersection of any functions, set them equal to each other:
x4-2 = x2
x4-x2-2 = 0
factoring this yields:
(x2-2)(x2+1)=0
by the zero identity, each of these factors can be set to zero to yield valid solutions:
x2-2 = 0
x2 = 2
x = +/- sqrt(2)
the second factor, when set equal to zero, yields an imaginary-valued solution. On a normal Cartesian coordinate plane, this means nothing. I assume you are only looking for points of intersection on a normal, real-valued Cartesian coordinate system (the normal x-y coordinate plane used in most elementary math classes).
So, these two functions intersect at x = -sqrt(2) and x = sqrt(2)
What else can I help you with?
Solve x2 equals 64?
x2 = 6482 = 64x = 8
X2 plus 7 equals 36?
if x2 + 7 = 37, then x2 = 29 and x = ±√29
Find a counterexample if x equals -5 then X2 equals 25?
In ordinary mathematics, assuming that x = X and that X2 denotes x2 or x-squared, there cannot be a counterexample since the statement is TRUE. However, there are two assumptions made that could be false and so could give rise to counterexamples. 1. x is not the same as X. If, for example X = 4x then X = -20 so that X2 = 400. 2a. X2 is not X2 but X times 2. In that case X2 = -10. 2b. X2 is x2 modulo 7, for example. Then X2 = 4.
What is x2 equals 9x-20?
x2=9x-20 x2-9x+20=0 Factor: (x-4)(x-5)=0 x={4,5}
X2 equals 11x - 10?
x2 = 11x - 10 ∴ x2 - 11x + 10 = 0 ∴ (x - 10)(x - 1) = 0 ∴ x ∈ {1, 10}
What are the points of intersection of y equals x squared plus 2x plus 2 with y equals x cubed plus 2?
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)
What are the coordinates for x2 plus y2 equals 185 and x plus y equals 17?
We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.
What are the points of intersection of the curves y equals x2 -7x plus 8 and y equals -x2 plus 9x -6 showing work?
If: y = x2-7x+8 and y = -x2+9x-6 Then: x2-7x+8 = -x2+9x-6 So: 2x2-16x+14 = 0 => x2-8x+7 = 0 Therefore: x = 1 and x = 7 By substitution: x =1, y = 2 and x = 7, y = 8 Points of intersection: (1, 2) and (7, 8)
How many points of intersection are the between x2 and x2 plus 5?
There are none. Those two curves do not intersect.
How many points does the circle x2 plus y2 equals 36 and the line y equals x-10 have in common?
X2 + Y2 = 36Y = (+/-) sqrt(36 - X2)=====================Radius = 6 (the circle is nicely oriented )andY = X - 10zeroing out variables in turnY = (0) - 10Y = - 10=======================With a radius of only six for the circle and the line with these intersection points we can say with some confidence that this circle has no points in common with this line.
What is the point of intersection of y equals 2x-x2 and y equals 0?
y = 2x - x2y = 0Since both quantities on the right side are equal to 'y', they're equal to each other.2x - x2 = 0x (2 - x) = 0x = 0and2 - x = 0x = 2The two points of intersection are (0, 0) and (2, 0) .
What is the point of intersection of the parabolas y equals x2 plus 20x plus 100 and y equals x2 -20x plus 100 showing work?
If: y = x2+20x+100 and x2-20x+100 Then: x2+20x+100 = x2-20x+100 So: 40x = 0 => x = 0 When x = 0 then y = 100 Therefore point of intersection: (0, 100)
How do you find the point of intersection between x2 plus y2 equals 5 and x-y equals 1?
x-y = 1 => x = y+1 x2+y2 = 5 => (y+1)(x+1)+y2 = 5 2y2+2y-4 = 0 y = -2 or y = 1 So the points of intersection are: (-1, -2) and (2, 1)
What are the points of intersection between the equations of 2x plus y equals 5 and x2 -y2 equals 3?
It works out that the points of intersection between the equations of 2x+5 = 5 and x^2 -y^2 = 3 are at: (14/3, -13/3) and (2, 1)
What is if any are the points of intersection of the parabolas y equals x2 -4x plus 8 and y equals 8x -x2 -14 showing work with explanations?
If: y = x2-4x+8 and y = 8x-x2-14 Then: x2-4x+8 = 8x-x2-14 So: 2x2-12x+22 = 0 Discriminant: 122-(4*2*22) = -32 Because the discriminant is less than 0 there is no actual contact between the given parabolas
What are the points of intersection between the line x -y equals 2 and the curve x2 -4y2 equals 5?
Equations: x -y = 2 and x^2 -4y^2 = 5 By combining the equations into a single quadratic equation in terms of y and solving it: y = 1/3 or y = 1 By means of substitution the points of intersection are at: (7/3, 1/3) and (3, 1)
What are the points of intersection of the line x -2 equals 2 with x2 -4y2 equals 5?
x - 2 = 2 → x = 4 → x² - 4y² = 5 → 4² - 4y² = 5 → 4y² = 16 - 5 → 4y² = 11 → y² = 11/4 → y = ± √(11/4) → The points of intersection of x - 2 = 2 with x² - 4y² = 5 are (4, -√(11/4)) ≈ 4, -1.658) and (4, √(11/4)) ≈ 4, 1.658)
