Find the points of contact of the quadritic curve y=x^2+5x-3 and the line 2y=3x-2.
2006-06-14 00:50:39 · 5 answers · asked by xavier h 1 in Science & Mathematics ➔ Mathematics
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2006-06-14 01:00:37 · update #1
y = x^2 + 5x - 3
2y=3x - 2
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2y = 2x^2 + 10x - 6
2y = 3x - 2
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0 = 2x^2 + 7x - 4
=>2x^2 + 7x - 4 =0
=>2x^2 + 8x - x - 4 = 0
=>2x(x + 4) -1(x+4) = 0
=> (2x - 1)(x + 4) = 0
therefore 2x - 1 = 0 => x = 1/2 or
x + 4 = 0 or x = -4
x = - 4 or 1/2
got it?
2006-06-14 01:13:56 · answer #1 · answered by Anonymous · 0⤊ 0⤋
y = x^2 + 5x - 3
2y = 3x - 2
2y = 3x - 2
y = (3/2)x - 1
(3/2)x - 1 = x^2 + 5x - 3
Multiply everything by 2
3x - 2 = 2x^2 + 10x - 6
2x^2 + 7x - 4 = 0
x = (-b ± sqrt(b^2 - 4ac))/2a
x = (-7 ± sqrt(7^2 - 4(2)(-4)))/(2(2))
x = (-7 ± sqrt(49 + 32))/4
x = (-7 ± sqrt(81))/4
x = (-7 ± 9)/4
x = (-16/4) or (2/4)
x = -4 or (1/2)
y = (3/2)x - 1
y = (3/2)(-4) - 1
y = -6 - 1
y = -7
or
y = (3/2)(1/2) - 1
y = (3/4) - 1
y = (3/4) - (4/4)
y = (-1/4)
ANS : (-4,-7) and (.5,-.25)
2006-06-14 08:18:23 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
x= - 4 ; y = -7
or
x = 1/2; y= -1/4
2006-06-14 00:57:16 · answer #3 · answered by à¹? (¯`v´¯)iChAi à¹? 2 · 0⤊ 0⤋
2y=3x-2 so
y=1.5x-1
solve
1.5x-1=x^2+5x-3
0=x^2+3.5x-2
0=(x+4)*(x-0.5)
x=-4 or x=0.5
2006-06-14 00:57:54 · answer #4 · answered by Anonymous · 0⤊ 0⤋
(.5,-.25) and (-4,-7)
2006-06-14 00:54:43 · answer #5 · answered by wIsEjEd1crack 1 · 0⤊ 0⤋