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1. 1. Solve the quadratic equation : 2x^2 - 3x +3 = 0
Use formula: x = [-b + or - square root(b^2 - 4*a*c)]/2a
x = [+3 + or - square root(9 - 24)]/4
x = [+ 3 + or - i*root(15)]/4 where i = square root(-1)
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2. Write as a single fraction with the denominator in factored form: 20[(2/(x+1)) - (3/x)]
lowest Common Denominator is x(x + 1)
20[2x -3(x + 1)]/[x(x + 1)]
20(-x - 3)/[x(x + 1)]
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3. sin(x) + tan(x)
sin(x) + sin(x)/cos(x)
[sin(x)*cos(x) + sin(x)]/cos(x)
sin(x)*[(cos(x) + 1]/cos(x)
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4. 4. Find the equation of the line that passes through the point [2,4] and is parallel to the line 2x + 3y -8 = 0
rewrite equation in standard form y = mx + c where m = slope and c the intercept on the y axis.
y = -2x/3 + 8/3
slope is -2/3 and intercept on Y axis is +8/3
Parallel line will have same slope of -2/3
substituting coordinates of point
4 = (-2/3)*2 + c>>>>>>>>>>>>>>>>> 4 = -4/3 + c
c = 16/3
equation is y = (-2/3)*x + 16/3 or 3y = -2x + 16
Note that a line perpendicular would have a slope of (+ 3/2)
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5. Find the equation of the line that is perpendicular to the line 2x + 3y - 8 = 0 at the point [1,2]
y = (-2/3)x + 8 Line perp has slope (+3/2)
New line: y = (3/2)x +c
Substitute for x and y
2 = (3/2)*1 + c >>>>>>>>>>>>>>>>>>> c = 2 - 3/2 = 1/2
c = +1/2
Equation y = (3/2)x + 1/2 or 2y = 3x + 1
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6. The line with slope 5 that passes through the point (-1, 3) intersects the x-axis at a point, what are the coordinates of this point?
y = 5x + c
Substitute given coords
3 = (-1)*5 + c
c = 8
equation y = 5x + 8
at intercept with X axis y = 0 so 5x = -8
x = -8/5
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Use formula: x = [-b + or - square root(b^2 - 4*a*c)]/2a
x = [+3 + or - square root(9 - 24)]/4
x = [+ 3 + or - i*root(15)]/4 where i = square root(-1)
--------------------------------------…
2. Write as a single fraction with the denominator in factored form: 20[(2/(x+1)) - (3/x)]
lowest Common Denominator is x(x + 1)
20[2x -3(x + 1)]/[x(x + 1)]
20(-x - 3)/[x(x + 1)]
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3. sin(x) + tan(x)
sin(x) + sin(x)/cos(x)
[sin(x)*cos(x) + sin(x)]/cos(x)
sin(x)*[(cos(x) + 1]/cos(x)
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4. 4. Find the equation of the line that passes through the point [2,4] and is parallel to the line 2x + 3y -8 = 0
rewrite equation in standard form y = mx + c where m = slope and c the intercept on the y axis.
y = -2x/3 + 8/3
slope is -2/3 and intercept on Y axis is +8/3
Parallel line will have same slope of -2/3
substituting coordinates of point
4 = (-2/3)*2 + c>>>>>>>>>>>>>>>>> 4 = -4/3 + c
c = 16/3
equation is y = (-2/3)*x + 16/3 or 3y = -2x + 16
Note that a line perpendicular would have a slope of (+ 3/2)
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5. Find the equation of the line that is perpendicular to the line 2x + 3y - 8 = 0 at the point [1,2]
y = (-2/3)x + 8 Line perp has slope (+3/2)
New line: y = (3/2)x +c
Substitute for x and y
2 = (3/2)*1 + c >>>>>>>>>>>>>>>>>>> c = 2 - 3/2 = 1/2
c = +1/2
Equation y = (3/2)x + 1/2 or 2y = 3x + 1
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6. The line with slope 5 that passes through the point (-1, 3) intersects the x-axis at a point, what are the coordinates of this point?
y = 5x + c
Substitute given coords
3 = (-1)*5 + c
c = 8
equation y = 5x + 8
at intercept with X axis y = 0 so 5x = -8
x = -8/5
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