關於中二MATHS題!(十五分)

題1: Find the coordiantes of the point of intersection of the parabola y=x2
and the line y=2x-1

y=x^2___________(1)
y=2x-1__________(2)
put(2)into(1),
2x-1=x2
x2-2x+1=0
(x-1)^2=0
x=1
put x=1 into (2)
y=2(1)-1
=1
the coordinates=(1,1)

題2: Find the x-intercepts of the parabola y=x2-5x+3

put y=0 into y=x2-5x+3
0=x2-5x+3
x2-5x+25/4=13/4
(x-5/2)^2=13/4
x-5/2=(√13)/2
x=(5√13)/2
x-intercepts of the parabola =(5√13)/2

題3: Find the x-intercept of the parabola y=x2-2(√2)x+2

put y=0 into y=x2-2(√2)x+2
0=x2-2√2x+2
(x-√2)^2=0
x=√2
x-intercepts of the parabola =√2
題4: Find the coordinates of the points of intersection of the parabola
y=x2 and y=x+3

y=x2__________________(1)
y=x+3_________________(2)
put (2) into (1),
x2=x+3
x2-x=3
x2-x+1/4=13/4
(x-1/2)^2=13)/4
x-1/2=(√13)/2
x=(1√13)/2
put x=(1√13)/2 into y=x+3
y=(7√13)/2
the coordinates=[(1√13)/2 ,(7√13)/2]
題5: Find the corodinates of the points of intersection of the parabola
y=x2-ax and the line y=bx-ab
y=x2_____________(1)
y=bx-ab __________(2)
put (2) into(1),
x2-ax=bx-ab
x2-(a+b)x+ab=0
(x-a)(x-b)=0
x=a or x=b
put x=a into(2),
y=ab-ab
=0
put x=b into(2),
y=b2-ab
the corodinates =(a,0) or(b,b2-ab)

(1)
y= 2x - 1 -----(1)
y = x^2 ------(2)
sub Eq(1) into Eq(2):
x^2 - 2x + 1 = 0
(x -1)^2 =0
when x = 1, y = 2(1) - 1 = 1
The intersection coordinates are (1,1)
(2)
y = x^2 - 5x + 3
the curve cuts x - axis which is the x-intercept
set y = 0, x^2 - 5x + 3 = 0
x = {5 + √[5^2 - 4(1)(3)]} /2 = 2.5 + 0.5√13
x = {5 - √[5^2 - 4(1)(3)]} /2 = 2.5 - 0.5√13
The x intercepts are (2.5 + 0.5√13, 0) and (2.5 - 0.5√13, 0)
(3)
the curve cuts x-axis which are the x intercepts
set y = 0, x^2 - 2√2x + 2 = 0
(x - √2)^2 = 0
x = √2
The x - intercept is ( √2, 0)
(4)
y = x + 3 -------(1)
y = x^2 -------(2)
sub Eq(1) into Eq(2):
x^2 = x + 3
x^2 - x - 3 = 0
x = {1 + √[(1)^2 - 4(1)(-3)]}/2 = 0.5 + 0.5√13
x = {1 - √[(1)^2 - 4(1)(-3)]}/2 = 0.5 - 0.5√13
The intersections are (0.5 + 0.5√13, 0) and (0.5 - 0.5√13, 0)
(5)
y = x^2 - ax ------(1)
y = bx - ab ------(2)
sub Eq(1) into Eq(2):
x^2 - ax = bx - ab
x^2 - (a+b)x + ab = 0
(x - a)(x - b) = 0
x = a, or x= b
when x = a, y = a^2 - a^2 = 0
when x = b, y = b^2 - ab
The intersection coordinates are (0, a) and (b, b^2 - ab)