Find the center-radius form of the equation of the circle having the points (3,−1) and (−5,3) as endpoints of a diameter.
Solution:
Center: midpoint of (3,−1) and (−5,3): (3+(2−5),−1+32 ) = (−1,1)
Radius: √
(3−(−1))2+(−5−3)2
2 =
√
42+(−8)2
2 =
√
16+64
2 =
√
80
2 =
4√5 2 =2
√
5
Find the center-radius form of the equation of the circle having the points (3,−1) and (−5,3) as endpoints of a diameter. Solution:
Center: midpoint of (3,−1) and (−5,3):
(3+(2−5),−1+32 ) = (−1,1)
Radius: √
(3−(−1))2+(−5−3)2
2 =
√
42+(−8)2
2 =
√
16+64
2 =
√
80
2 =
4√5 2 =2
√
5
Find the center-radius form of the equation of the circle having the points (3,−1) and (−5,3) as endpoints of a diameter. Solution:
Center: midpoint of (3,−1) and (−5,3): (3+(2−5),−1+32 )
= (−1,1)
Radius: √
(3−(−1))2+(−5−3)2
2 =
√
42+(−8)2
2 =
√
16+64
2 =
√
80
2 =
4√5 2 =2
√
5
Find the center-radius form of the equation of the circle having the points (3,−1) and (−5,3) as endpoints of a diameter. Solution:
Center: midpoint of (3,−1) and (−5,3): (3+(2−5),−1+32 ) = (−1,1)
Radius: √
(3−(−1))2+(−5−3)2
2 =
√
42+(−8)2
2 =
√
16+64
2 =
√
80
2 =
4√5 2 =2
√
5
Find the center-radius form of the equation of the circle having the points (3,−1) and (−5,3) as endpoints of a diameter. Solution:
Center: midpoint of (3,−1) and (−5,3): (3+(2−5),−1+32 ) = (−1,1)
Radius: √
(3−(−1))2+(−5−3)2 2
= √
42+(−8)2
2 =
√
16+64
2 =
√
80
2 =
4√5 2 =2
√
5
Find the center-radius form of the equation of the circle having the points (3,−1) and (−5,3) as endpoints of a diameter. Solution:
Center: midpoint of (3,−1) and (−5,3): (3+(2−5),−1+32 ) = (−1,1)
Radius: √
(3−(−1))2+(−5−3)2
2 =
√
42+(−8)2 2
= √
16+64
2 =
√
80
2 =
4√5 2 =2
√
5
Find the center-radius form of the equation of the circle having the points (3,−1) and (−5,3) as endpoints of a diameter. Solution:
Center: midpoint of (3,−1) and (−5,3): (3+(2−5),−1+32 ) = (−1,1)
Radius: √
(3−(−1))2+(−5−3)2
2 =
√
42+(−8)2
2 =
√
16+64 2
= √
80
2 =
4√5 2 =2
√
5
Find the center-radius form of the equation of the circle having the points (3,−1) and (−5,3) as endpoints of a diameter. Solution:
Center: midpoint of (3,−1) and (−5,3): (3+(2−5),−1+32 ) = (−1,1)
Radius: √
(3−(−1))2+(−5−3)2
2 =
√
42+(−8)2
2 =
√
16+64
2 =
√
80 2
=
4√5 2 =2
√
5
Find the center-radius form of the equation of the circle having the points (3,−1) and (−5,3) as endpoints of a diameter. Solution:
Center: midpoint of (3,−1) and (−5,3): (3+(2−5),−1+32 ) = (−1,1)
Radius: √
(3−(−1))2+(−5−3)2
2 =
√
42+(−8)2
2 =
√
16+64
2 =
√
80
2 =
4√5 2
=2√5
Find the center-radius form of the equation of the circle having the points (3,−1) and (−5,3) as endpoints of a diameter. Solution:
Center: midpoint of (3,−1) and (−5,3): (3+(2−5),−1+32 ) = (−1,1)
Radius: √
(3−(−1))2+(−5−3)2
2 =
√
42+(−8)2
2 =
√
16+64
2 =
√
80
2 =
4√5 2 =2
√
5
Find the center-radius form of the equation of the circle having the points (3,−1) and (−5,3) as endpoints of a diameter. Solution:
Center: midpoint of (3,−1) and (−5,3): (3+(2−5),−1+32 ) = (−1,1)
Radius: √
(3−(−1))2+(−5−3)2
2 =
√
42+(−8)2
2 =
√
16+64
2 =
√
80
2 =
4√5 2 =2
√
5
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
1. determine the x- and y-intercepts of the graphs of the given equations
Solution: line:
y-int: 4(0) +y= 3
⇒y= 3
x-int: 4x+ 0 = 3⇒x= 34
parabola: y-int:
y= 4(0)2−8(0)−5⇒y=−5 x-int: 0 = 4x2−8x−5
⇒0 = (2x−5)(2x+ 1) ⇒x= 52 orx=−1
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
1. determine the x- and y-intercepts of the graphs of the given equations
Solution: line:
y-int: 4(0) +y= 3⇒y= 3
x-int: 4x+ 0 = 3⇒x= 34
parabola: y-int:
y= 4(0)2−8(0)−5⇒y=−5 x-int: 0 = 4x2−8x−5
⇒0 = (2x−5)(2x+ 1) ⇒x= 52 orx=−1
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
1. determine the x- and y-intercepts of the graphs of the given equations
Solution: line:
y-int: 4(0) +y= 3⇒y= 3
x-int: 4x+ 0 = 3
⇒x= 34
parabola: y-int:
y= 4(0)2−8(0)−5⇒y=−5 x-int: 0 = 4x2−8x−5
⇒0 = (2x−5)(2x+ 1) ⇒x= 52 orx=−1
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
1. determine the x- and y-intercepts of the graphs of the given equations
Solution: line:
y-int: 4(0) +y= 3⇒y= 3
x-int: 4x+ 0 = 3⇒x= 34
parabola: y-int:
y= 4(0)2−8(0)−5⇒y=−5 x-int: 0 = 4x2−8x−5
⇒0 = (2x−5)(2x+ 1) ⇒x= 52 orx=−1
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
1. determine the x- and y-intercepts of the graphs of the given equations
Solution: line:
y-int: 4(0) +y= 3⇒y= 3
x-int: 4x+ 0 = 3⇒x= 34
parabola: y-int:
y= 4(0)2−8(0)−5
⇒y=−5 x-int: 0 = 4x2−8x−5
⇒0 = (2x−5)(2x+ 1) ⇒x= 52 orx=−1
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
1. determine the x- and y-intercepts of the graphs of the given equations
Solution: line:
y-int: 4(0) +y= 3⇒y= 3
x-int: 4x+ 0 = 3⇒x= 34
parabola: y-int:
y= 4(0)2−8(0)−5⇒y=−5
x-int: 0 = 4x2−8x−5 ⇒0 = (2x−5)(2x+ 1) ⇒x= 52 orx=−1
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
1. determine the x- and y-intercepts of the graphs of the given equations
Solution: line:
y-int: 4(0) +y= 3⇒y= 3
x-int: 4x+ 0 = 3⇒x= 34
parabola: y-int:
y= 4(0)2−8(0)−5⇒y=−5 x-int: 0 = 4x2−8x−5
⇒0 = (2x−5)(2x+ 1) ⇒x= 52 orx=−1
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
1. determine the x- and y-intercepts of the graphs of the given equations
Solution: line:
y-int: 4(0) +y= 3⇒y= 3
x-int: 4x+ 0 = 3⇒x= 34
parabola: y-int:
y= 4(0)2−8(0)−5⇒y=−5 x-int: 0 = 4x2−8x−5
⇒0 = (2x−5)(2x+ 1)
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
1. determine the x- and y-intercepts of the graphs of the given equations
Solution: line:
y-int: 4(0) +y= 3⇒y= 3
x-int: 4x+ 0 = 3⇒x= 34
parabola: y-int:
y= 4(0)2−8(0)−5⇒y=−5 x-int: 0 = 4x2−8x−5
⇒0 = (2x−5)(2x+ 1) ⇒x= 52 orx=−1
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
2. find the vertex of the parabola Solution:
−b
2a,
4ac−b2 4a
=
−(−8)
2(4) ,
4(4)(−5)−(−8)2
4(4)
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
2. find the vertex of the parabola Solution:
−b
2a,
4ac−b2 4a
=
−(−8)
2(4) ,
4(4)(−5)−(−8)2
4(4)
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
2. find the vertex of the parabola Solution:
−b
2a,
4ac−b2 4a
=
−(−8)
2(4) ,
4(4)(−5)−(−8)2
4(4)
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
3. solve algebraically for the point/s of intersection of the graphs of the given equations
Solutions:
Via substitution,
4x2−8x−5 =−4x+ 3
4x2−4x−8 = 0 4(x2−x−2) = 0 4(x−2)(x+ 1) = 0
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
3. solve algebraically for the point/s of intersection of the graphs of the given equations
Solutions:
Via substitution,
4x2−8x−5 =−4x+ 3 4x2−4x−8 = 0
4(x2−x−2) = 0 4(x−2)(x+ 1) = 0
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
3. solve algebraically for the point/s of intersection of the graphs of the given equations
Solutions:
Via substitution,
4x2−8x−5 =−4x+ 3 4x2−4x−8 = 0
4(x2−x−2) = 0
4(x−2)(x+ 1) = 0
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
3. solve algebraically for the point/s of intersection of the graphs of the given equations
Solutions:
Via substitution,
4x2−8x−5 =−4x+ 3 4x2−4x−8 = 0
4(x2−x−2) = 0 4(x−2)(x+ 1) = 0
Given the parabola with equationy= 4x2−8x−5 and the line with equation4x+y= 3.
3. solve algebraically for the point/s of intersection of the graphs of the given equations
Solutions:
Via substitution,
4x2−8x−5 =−4x+ 3 4x2−4x−8 = 0
4(x2−x−2) = 0 4(x−2)(x+ 1) = 0
ifx= 2,
Given the parabola with equationy= 4x2−8x−5 and the line with equation4x+y= 3.
3. solve algebraically for the point/s of intersection of the graphs of the given equations
Solutions:
Via substitution,
4x2−8x−5 =−4x+ 3 4x2−4x−8 = 0
4(x2−x−2) = 0 4(x−2)(x+ 1) = 0
ifx= 2, y=−5
Given the parabola with equationy= 4x2−8x−5 and the line with equation4x+y= 3.
3. solve algebraically for the point/s of intersection of the graphs of the given equations
Solutions:
Via substitution,
4x2−8x−5 =−4x+ 3 4x2−4x−8 = 0
4(x2−x−2) = 0 4(x−2)(x+ 1) = 0
ifx= 2, y=−5 ifx=−1,
Given the parabola with equationy= 4x2−8x−5 and the line with equation4x+y= 3.
3. solve algebraically for the point/s of intersection of the graphs of the given equations
Solutions:
Via substitution,
4x2−8x−5 =−4x+ 3 4x2−4x−8 = 0
4(x2−x−2) = 0 4(x−2)(x+ 1) = 0
ifx= 2, y=−5 ifx=−1, y= 7
Given the parabola with equationy= 4x2−8x−5 and the line with equation 4x+y= 3.
3. solve algebraically for the point/s of intersection of the graphs of the given equations
Solutions:
Via substitution,
4x2−8x−5 =−4x+ 3 4x2−4x−8 = 0
4(x2−x−2) = 0 4(x−2)(x+ 1) = 0
The graphs ofy= 4x2−8x−5 and 4x+y= 3.
−2 2 4 6 8
−10
−8
−6
−4
−2 2 4 6 8 (−1,7)
Give the solution set of the following:
1. 3x−8
x2−3x−10 +
2x−3 x−5 =
x+ 4 x+ 2 Solution:
3x−8 (x−5)(x+ 2) +
2x−3 x−5 =
x+ 4
x+ 2 factor the denominators (3x−8) + (2x−3)(x+ 2) = (x+ 4)(x−5) multiply the LCD
(3x−8) + (2x2+x−6) = (x2−x−20)
x2+ 5x+ 6 = 0 additive property
(x+ 2)(x+ 3) = 0 factor the LHS
Give the solution set of the following:
1. 3x−8
x2−3x−10 +
2x−3 x−5 =
x+ 4 x+ 2 Solution:
3x−8 (x−5)(x+ 2) +
2x−3 x−5 =
x+ 4
x+ 2 factor the denominators
(3x−8) + (2x−3)(x+ 2) = (x+ 4)(x−5) multiply the LCD (3x−8) + (2x2+x−6) = (x2−x−20)
x2+ 5x+ 6 = 0 additive property
(x+ 2)(x+ 3) = 0 factor the LHS
Give the solution set of the following:
1. 3x−8
x2−3x−10 +
2x−3 x−5 =
x+ 4 x+ 2 Solution:
3x−8 (x−5)(x+ 2) +
2x−3 x−5 =
x+ 4
x+ 2 factor the denominators (3x−8) + (2x−3)(x+ 2) = (x+ 4)(x−5) multiply the LCD
(3x−8) + (2x2+x−6) = (x2−x−20)
x2+ 5x+ 6 = 0 additive property
(x+ 2)(x+ 3) = 0 factor the LHS
Give the solution set of the following:
1. 3x−8
x2−3x−10 +
2x−3 x−5 =
x+ 4 x+ 2 Solution:
3x−8 (x−5)(x+ 2) +
2x−3 x−5 =
x+ 4
x+ 2 factor the denominators (3x−8) + (2x−3)(x+ 2) = (x+ 4)(x−5) multiply the LCD
(3x−8) + (2x2+x−6) = (x2−x−20)
x2+ 5x+ 6 = 0 additive property
(x+ 2)(x+ 3) = 0 factor the LHS
Give the solution set of the following:
1. 3x−8
x2−3x−10 +
2x−3 x−5 =
x+ 4 x+ 2 Solution:
3x−8 (x−5)(x+ 2) +
2x−3 x−5 =
x+ 4
x+ 2 factor the denominators (3x−8) + (2x−3)(x+ 2) = (x+ 4)(x−5) multiply the LCD
(3x−8) + (2x2+x−6) = (x2−x−20)
x2+ 5x+ 6 = 0 additive property
(x+ 2)(x+ 3) = 0 factor the LHS
Give the solution set of the following:
1. 3x−8
x2−3x−10 +
2x−3 x−5 =
x+ 4 x+ 2 Solution:
3x−8 (x−5)(x+ 2) +
2x−3 x−5 =
x+ 4
x+ 2 factor the denominators (3x−8) + (2x−3)(x+ 2) = (x+ 4)(x−5) multiply the LCD
(3x−8) + (2x2+x−6) = (x2−x−20)
x2+ 5x+ 6 = 0 additive property
(x+ 2)(x+ 3) = 0 factor the LHS
Give the solution set of the following:
1. 3x−8
x2−3x−10 +
2x−3 x−5 =
x+ 4 x+ 2 Solution:
3x−8 (x−5)(x+ 2) +
2x−3 x−5 =
x+ 4
x+ 2 factor the denominators (3x−8) + (2x−3)(x+ 2) = (x+ 4)(x−5) multiply the LCD
(3x−8) + (2x2+x−6) = (x2−x−20)
x2+ 5x+ 6 = 0 additive property
(x+ 2)(x+ 3) = 0 factor the LHS
x=−2 orx=−3
Give the solution set of the following:
1. 3x−8
x2−3x−10 +
2x−3 x−5 =
x+ 4 x+ 2 Solution:
3x−8 (x−5)(x+ 2) +
2x−3 x−5 =
x+ 4
x+ 2 factor the denominators (3x−8) + (2x−3)(x+ 2) = (x+ 4)(x−5) multiply the LCD
(3x−8) + (2x2+x−6) = (x2−x−20)
x2+ 5x+ 6 = 0 additive property
(x+ 2)(x+ 3) = 0 factor the LHS
Give the solution set of the following:
2. x23 −2x 1
3 −3 = 0
Solution:
w2−2w−3 = 0 quadratic form when w=x13
(w−3)(w+ 1) = 0 factoring
w= 3 orw=−1
x13 = 3
(x13)3 = (3)3
x= 27
x13 =−1
(x13)3 = (−1)3
x=−1
Checking: (−1)23 −2(−1)
1 3 −
3 = 1 + 2−3 = 0 2723 −2(27)
1 3 −3 =
Give the solution set of the following:
2. x23 −2x 1
3 −3 = 0
Solution:
w2−2w−3 = 0 quadratic form when w=x13
(w−3)(w+ 1) = 0 factoring
w= 3 orw=−1
x13 = 3
(x13)3 = (3)3
x= 27
x13 =−1
(x13)3 = (−1)3
x=−1
Checking: (−1)23 −2(−1)
1 3 −
3 = 1 + 2−3 = 0 2723 −2(27)
1 3 −3 =
Give the solution set of the following:
2. x23 −2x 1
3 −3 = 0
Solution:
w2−2w−3 = 0 quadratic form when w=x13
(w−3)(w+ 1) = 0 factoring
w= 3 orw=−1
x13 = 3
(x13)3 = (3)3
x= 27
x13 =−1
(x13)3 = (−1)3
x=−1
Checking: (−1)23 −2(−1)
1 3 −
3 = 1 + 2−3 = 0 2723 −2(27)
1 3 −3 =
Give the solution set of the following:
2. x23 −2x 1
3 −3 = 0
Solution:
w2−2w−3 = 0 quadratic form when w=x13
(w−3)(w+ 1) = 0 factoring
w= 3 orw=−1
x13 = 3
(x13)3 = (3)3
x= 27
x13 =−1
(x13)3 = (−1)3
x=−1
Checking: (−1)23 −2(−1)
1 3 −
3 = 1 + 2−3 = 0 2723 −2(27)
1 3 −3 =
Give the solution set of the following:
2. x23 −2x 1
3 −3 = 0
Solution:
w2−2w−3 = 0 quadratic form when w=x13
(w−3)(w+ 1) = 0 factoring
w= 3 orw=−1
x13 = 3
(x13)3 = (3)3
x= 27
x13 =−1
(x13)3 = (−1)3
x=−1
Checking: (−1)23 −2(−1)
1 3 −
3 = 1 + 2−3 = 0 2723 −2(27)
1 3 −3 =
Give the solution set of the following:
2. x23 −2x 1
3 −3 = 0
Solution:
w2−2w−3 = 0 quadratic form when w=x13
(w−3)(w+ 1) = 0 factoring
w= 3 orw=−1
x13 = 3
(x13)3 = (3)3
x= 27
x13 =−1
(x13)3 = (−1)3
x=−1
Checking: (−1)23 −2(−1)
1 3 −
3 = 1 + 2−3 = 0 2723 −2(27)
1 3 −3 =
Give the solution set of the following:
2. x23 −2x 1
3 −3 = 0
Solution:
w2−2w−3 = 0 quadratic form when w=x13
(w−3)(w+ 1) = 0 factoring
w= 3 orw=−1
x13 = 3
(x13)3 = (3)3
x= 27
x13 =−1
(x13)3 = (−1)3
x=−1
Checking: (−1)23 −2(−1)
1 3 −
3 = 1 + 2−3 = 0 2723 −2(27)
1 3 −3 =
Give the solution set of the following:
2. x23 −2x 1
3 −3 = 0
Solution:
w2−2w−3 = 0 quadratic form when w=x13
(w−3)(w+ 1) = 0 factoring
w= 3 orw=−1
x13 = 3
(x13)3 = (3)3
x= 27
x13 =−1
(x13)3 = (−1)3
x=−1
Checking: (−1)23 −2(−1)
1 3 −
3 = 1 + 2−3 = 0 2723 −2(27)
1 3 −3 =
Give the solution set of the following:
2. x23 −2x 1
3 −3 = 0
Solution:
w2−2w−3 = 0 quadratic form when w=x13
(w−3)(w+ 1) = 0 factoring
w= 3 orw=−1
x13 = 3
(x13)3 = (3)3
x= 27
x13 =−1
(x13)3 = (−1)3
x=−1
Checking: (−1)23 −2(−1)
1 3 −
3 = 1 + 2−3 = 0 2723 −2(27)
1 3 −3 =
Give the solution set of the following:
2. x23 −2x 1
3 −3 = 0
Solution:
w2−2w−3 = 0 quadratic form when w=x13
(w−3)(w+ 1) = 0 factoring
w= 3 orw=−1
x13 = 3
(x13)3 = (3)3
x= 27
x13 =−1
(x13)3 = (−1)3
x=−1
Checking: (−1)23 −2(−1)
1 3 −
3 = 1 + 2−3 = 0 2723 −2(27)
1 3 −3 =
Give the solution set of the following:
2. x23 −2x 1
3 −3 = 0
Solution:
w2−2w−3 = 0 quadratic form when w=x13
(w−3)(w+ 1) = 0 factoring
w= 3 orw=−1
x13 = 3
(x13)3 = (3)3
x= 27
x13 =−1
(x13)3 = (−1)3
x=−1
Checking: (−1)23 −2(−1)
1 3 −
3 = 1 + 2−3 = 0 2723 −2(27)
1 3 −3 =
9−2(3)−3 = 0
Give the solution set of the following:
2. x23 −2x 1
3 −3 = 0
Solution:
w2−2w−3 = 0 quadratic form when w=x13
(w−3)(w+ 1) = 0 factoring
w= 3 orw=−1
x13 = 3
(x13)3 = (3)3
x= 27
x13 =−1
(x13)3 = (−1)3
x=−1
Checking: (−1)23 −2(−1)
1 3 −
3 = 1 + 2−3 = 0 2723 −2(27)
1 3 −3 =
Give the solution set of the following:
3. px−√10−x= 2
Solution:
x−√10−x= 4 square both sides x−4 =√10−x additive property x2−8x+ 16 = 10−x square both sides x2−7x+ 6 = 0 additive property (x−6)(x−1) = 0 factoring
Checking: p
6−√10−6 =√6−2 = 2 p
1−√10−1 =√1−36= 2
Give the solution set of the following:
3. px−√10−x= 2 Solution:
x−√10−x= 4 square both sides
x−4 =√10−x additive property x2−8x+ 16 = 10−x square both sides x2−7x+ 6 = 0 additive property (x−6)(x−1) = 0 factoring
Checking: p
6−√10−6 =√6−2 = 2 p
1−√10−1 =√1−36= 2
Give the solution set of the following:
3. px−√10−x= 2 Solution:
x−√10−x= 4 square both sides x−4 =√10−x additive property
x2−8x+ 16 = 10−x square both sides x2−7x+ 6 = 0 additive property (x−6)(x−1) = 0 factoring
Checking: p
6−√10−6 =√6−2 = 2 p
1−√10−1 =√1−36= 2
Give the solution set of the following:
3. px−√10−x= 2 Solution:
x−√10−x= 4 square both sides x−4 =√10−x additive property x2−8x+ 16 = 10−x square both sides
x2−7x+ 6 = 0 additive property (x−6)(x−1) = 0 factoring
Checking: p
6−√10−6 =√6−2 = 2 p
1−√10−1 =√1−36= 2
Give the solution set of the following:
3. px−√10−x= 2 Solution:
x−√10−x= 4 square both sides x−4 =√10−x additive property x2−8x+ 16 = 10−x square both sides x2−7x+ 6 = 0 additive property
(x−6)(x−1) = 0 factoring
Checking: p
6−√10−6 =√6−2 = 2 p
1−√10−1 =√1−36= 2
Give the solution set of the following:
3. px−√10−x= 2 Solution:
x−√10−x= 4 square both sides x−4 =√10−x additive property x2−8x+ 16 = 10−x square both sides x2−7x+ 6 = 0 additive property (x−6)(x−1) = 0 factoring
Checking: p
6−√10−6 =√6−2 = 2 p
1−√10−1 =√1−36= 2
Give the solution set of the following:
3. px−√10−x= 2 Solution:
x−√10−x= 4 square both sides x−4 =√10−x additive property x2−8x+ 16 = 10−x square both sides x2−7x+ 6 = 0 additive property (x−6)(x−1) = 0 factoring
Checking: p
6−√10−6 =√6−2 = 2 p
1−√10−1 =√1−36= 2
Give the solution set of
x−2 2x−1
≥1
Solution:
x−2
2x−1 ≥1 or
x−2 2x−1 ≤ −1
x−2
2x−1 −1≥0 (x−2)−(2x−1)
2x−1 ≥0 −x−1
2x−1 ≥0 CN :−1,1
2
x−2
2x−1 + 1≤0 (x−2) + (2x−1)
2x−1 ≤0 3x−3 2x−1 ≤0 CN : 1,1
Give the solution set of
x−2 2x−1
≥1 Solution:
x−2
2x−1 ≥1 or
x−2 2x−1 ≤ −1
x−2
2x−1 −1≥0 (x−2)−(2x−1)
2x−1 ≥0 −x−1
2x−1 ≥0 CN :−1,1
2
x−2
2x−1 + 1≤0 (x−2) + (2x−1)
2x−1 ≤0 3x−3 2x−1 ≤0 CN : 1,1
Give the solution set of
x−2 2x−1
≥1 Solution:
x−2
2x−1 ≥1 or
x−2 2x−1 ≤ −1
x−2
2x−1 −1≥0
(x−2)−(2x−1) 2x−1 ≥0
−x−1 2x−1 ≥0 CN :−1,1
2
x−2
2x−1 + 1≤0 (x−2) + (2x−1)
2x−1 ≤0 3x−3 2x−1 ≤0 CN : 1,1
Give the solution set of
x−2 2x−1
≥1 Solution:
x−2
2x−1 ≥1 or
x−2 2x−1 ≤ −1
x−2
2x−1 −1≥0 (x−2)−(2x−1)
2x−1 ≥0
−x−1 2x−1 ≥0 CN :−1,1
2
x−2
2x−1 + 1≤0 (x−2) + (2x−1)
2x−1 ≤0 3x−3 2x−1 ≤0 CN : 1,1
Give the solution set of
x−2 2x−1
≥1 Solution:
x−2
2x−1 ≥1 or
x−2 2x−1 ≤ −1
x−2
2x−1 −1≥0 (x−2)−(2x−1)
2x−1 ≥0 −x−1
2x−1 ≥0
CN :−1,1 2
x−2
2x−1 + 1≤0 (x−2) + (2x−1)
2x−1 ≤0 3x−3 2x−1 ≤0 CN : 1,1
Give the solution set of
x−2 2x−1
≥1 Solution:
x−2
2x−1 ≥1 or
x−2 2x−1 ≤ −1
x−2
2x−1 −1≥0 (x−2)−(2x−1)
2x−1 ≥0 −x−1
2x−1 ≥0 CN :−1,1
2
x−2
2x−1 + 1≤0 (x−2) + (2x−1)
2x−1 ≤0 3x−3 2x−1 ≤0 CN : 1,1
Give the solution set of
x−2 2x−1
≥1 Solution:
x−2
2x−1 ≥1 or
x−2 2x−1 ≤ −1
x−2
2x−1 −1≥0 (x−2)−(2x−1)
2x−1 ≥0 −x−1
2x−1 ≥0 CN :−1,1
2
x−2
2x−1 + 1≤0
(x−2) + (2x−1) 2x−1 ≤0
3x−3 2x−1 ≤0 CN : 1,1
Give the solution set of
x−2 2x−1
≥1 Solution:
x−2
2x−1 ≥1 or
x−2 2x−1 ≤ −1
x−2
2x−1 −1≥0 (x−2)−(2x−1)
2x−1 ≥0 −x−1
2x−1 ≥0 CN :−1,1
2
x−2
2x−1 + 1≤0 (x−2) + (2x−1)
2x−1 ≤0
3x−3 2x−1 ≤0 CN : 1,1
Give the solution set of
x−2 2x−1
≥1 Solution:
x−2
2x−1 ≥1 or
x−2 2x−1 ≤ −1
x−2
2x−1 −1≥0 (x−2)−(2x−1)
2x−1 ≥0 −x−1
2x−1 ≥0 CN :−1,1
2
x−2
2x−1 + 1≤0 (x−2) + (2x−1)
2x−1 ≤0 3x−3 2x−1 ≤0
Give the solution set of
x−2 2x−1
≥1 Solution:
x−2
2x−1 ≥1 or
x−2 2x−1 ≤ −1
x−2
2x−1 −1≥0 (x−2)−(2x−1)
2x−1 ≥0 −x−1
2x−1 ≥0 CN :−1,1
2
x−2
2x−1 + 1≤0 (x−2) + (2x−1)
2x−1 ≤0 3x−3 2x−1 ≤0 CN : 1,1
Give the solution set of
x−2 2x−1
≥1 Solution(cont):
−x−1 2x−1 ≥0:
(−∞,−1) (−1,1 2) (
1 2,∞)
−x−1 + -
-2x−1 - - +
x−2
2x−1 - +
-3x−3 2x−1 ≤0:
(−∞,1 2) (
1
2,1) (1,∞)
3x−3 - - +
2x−1 - + +
x−2
2x−1 + - +
SS: [−1,12)S
Give the solution set of
x−2 2x−1
≥1 Solution(cont):
−x−1 2x−1 ≥0:
(−∞,−1) (−1,1 2) (
1 2,∞)
−x−1 + -
-2x−1 - - +
x−2
2x−1 - +
-3x−3 2x−1 ≤0:
(−∞,1 2) (
1
2,1) (1,∞)
3x−3 - - +
2x−1 - + +
x−2
2x−1 + - +
SS: [−1,12)S
Give the solution set of
x−2 2x−1
≥1 Solution(cont):
−x−1 2x−1 ≥0:
(−∞,−1) (−1,1 2) (
1 2,∞)
−x−1 + -
-2x−1 - - +
x−2
2x−1 - +
-3x−3 2x−1 ≤0:
(−∞,1 2) (
1
2,1) (1,∞)
3x−3 - - +
2x−1 - + +
x−2
2x−1 + - +
SS: [−1,12)S
Give the solution set of
x−2 2x−1
≥1 Solution(cont):
−x−1 2x−1 ≥0:
(−∞,−1) (−1,1 2) (
1 2,∞)
−x−1 + -
-2x−1 - - +
x−2
2x−1 - +
-3x−3 2x−1 ≤0:
(−∞,12) (12,1) (1,∞)
3x−3 - - +
2x−1 - + +
x−2
2x−1 + - +
SS: [−1,12)S
Give the solution set of
x−2 2x−1
≥1 Solution(cont):
−x−1 2x−1 ≥0:
(−∞,−1) (−1,1 2) (
1 2,∞)
−x−1 + -
-2x−1 - - +
x−2
2x−1 - +
-3x−3 2x−1 ≤0:
(−∞,12) (12,1) (1,∞)
3x−3 - - +
2x−1 - + +
x−2
2x−1 + - +
SS: [−1,12)S
Ifz varies directly asy and inversely as the square root of x, by how much shouldx be increased so that z is kept constant wheny is increased by 10%?
Solution:
z=k√y x
Increasingy by 10% means substituting y+ 0.1y fory. Increasingx by an amountmeans substituting x+forx.
Doing both, we have a new value for z: k√1.1y x+. k√(1.1y)
x+ = ky √
x (z is kept constant)
1.1
√
x+ =
1
√
x (divide both sides by
1
yk) x+
1.21 = x (take reciprocals then square)
Ifz varies directly asy and inversely as the square root of x, by how much shouldx be increased so that z is kept constant wheny is increased by 10%?
Solution:
z=k√y x
Increasingy by 10% means substituting y+ 0.1y fory. Increasingx by an amountmeans substituting x+forx.
Doing both, we have a new value for z: k√1.1y x+. k√(1.1y)
x+ = ky √
x (z is kept constant)
1.1
√
x+ =
1
√
x (divide both sides by
1
yk) x+
1.21 = x (take reciprocals then square)
Ifz varies directly asy and inversely as the square root of x, by how much shouldx be increased so that z is kept constant wheny is increased by 10%?
Solution:
z=k√y x
Increasingy by 10% means substituting y+ 0.1y fory.
Increasingx by an amountmeans substituting x+forx. Doing both, we have a new value for z: k√1.1y
x+. k√(1.1y)
x+ = ky √
x (z is kept constant)
1.1
√
x+ =
1
√
x (divide both sides by
1
yk) x+
1.21 = x (take reciprocals then square)
Ifz varies directly asy and inversely as the square root of x,by how much shouldx be increasedso that z is kept constant wheny is increased by 10%?
Solution:
z=k√y x
Increasingy by 10% means substituting y+ 0.1y fory. Increasingx by an amountmeans substitutingx+forx.
Doing both, we have a new value for z: k√1.1y x+. k√(1.1y)
x+ = ky √
x (z is kept constant)
1.1
√
x+ =
1
√
x (divide both sides by
1
yk) x+
1.21 = x (take reciprocals then square)
Ifz varies directly asy and inversely as the square root of x, by how much shouldx be increased so that z is kept constant wheny is increased by 10%?
Solution:
z=k√y x
Increasingy by 10% means substituting y+ 0.1y fory. Increasingx by an amountmeans substitutingx+forx.
Doing both, we have a new value for z: k√1.1y x+.
k√(1.1y)
x+ = ky √
x (z is kept constant)
1.1
√
x+ =
1
√
x (divide both sides by
1
yk) x+
1.21 = x (take reciprocals then square)
Ifz varies directly asy and inversely as the square root of x, by how much shouldx be increased so that z is kept constant wheny is increased by 10%?
Solution:
z=k√y x
Increasingy by 10% means substituting y+ 0.1y fory. Increasingx by an amountmeans substitutingx+forx.
Doing both, we have a new value for z: k√1.1y x+. k√(1.1y)
x+ = ky √
x (z is kept constant)
1.1
√
x+ =
1
√
x (divide both sides by
1
yk) x+
1.21 = x (take reciprocals then square)
Ifz varies directly asy and inversely as the square root of x, by how much shouldx be increased so that z is kept constant wheny is increased by 10%?
Solution:
z=k√y x
Increasingy by 10% means substituting y+ 0.1y fory. Increasingx by an amountmeans substitutingx+forx.
Doing both, we have a new value for z: k√1.1y x+. k√(1.1y)
x+ = ky √
x (z is kept constant)
1.1
√
x+ =
1
√
x (divide both sides by
1
yk)
x+
1.21 = x (take reciprocals then square)
Ifz varies directly asy and inversely as the square root of x, by how much shouldx be increased so that z is kept constant wheny is increased by 10%?
Solution:
z=k√y x
Increasingy by 10% means substituting y+ 0.1y fory. Increasingx by an amountmeans substitutingx+forx.
Doing both, we have a new value for z: k√1.1y x+. k√(1.1y)
x+ = ky √
x (z is kept constant)
1.1
√
x+ =
1
√
x (divide both sides by
1
yk) x+
1.21 = x (take reciprocals then square)
Ifz varies directly asy and inversely as the square root of x, by how much shouldx be increased so that z is kept constant wheny is increased by 10%?
Solution:
z=k√y x
Increasingy by 10% means substituting y+ 0.1y fory. Increasingx by an amountmeans substitutingx+forx.
Doing both, we have a new value for z: k√1.1y x+. k√(1.1y)
x+ = ky √
x (z is kept constant)
1.1
√
x+ =
1
√
x (divide both sides by
1
yk) x+
1.21 = x (take reciprocals then square)
x+ = 1.21x (multiplicative prop.) = 0.21x (additive prop.)
Ifz varies directly asy and inversely as the square root of x, by how much shouldx be increased so that z is kept constant wheny is increased by 10%?
Solution:
z=k√y x
Increasingy by 10% means substituting y+ 0.1y fory. Increasingx by an amountmeans substitutingx+forx.
Doing both, we have a new value for z: k√1.1y x+. k√(1.1y)
x+ = ky √
x (z is kept constant)
1.1
√
x+ =
1
√
x (divide both sides by
1
yk) x+
1.21 = x (take reciprocals then square)
A boat goes downstream a distance of 5 kilometers in 15 minutes but it takes 20 minutes for the return trip. Find the rate of the current.
Solution:
Let x be the boat’s speed, y be the current’s speed (in km. per min.)
x+y = 155 (1) x−y = 205 (2) Subtracting (2) from (1): 2y= 13 −1
4 = 1 12
y = 12 121
A boat goes downstream a distance of 5 kilometers in 15 minutes but it takes 20 minutes for the return trip. Find the rate of the current.
Solution:
Let x be the boat’s speed, y be the current’s speed (in km. per min.)
x+y = 155 (1) x−y = 205 (2) Subtracting (2) from (1): 2y= 13 −1
4 = 1 12
y = 12 121
A boat goes downstream a distance of 5 kilometers in 15 minutes but it takes 20 minutes for the return trip. Find the rate of the current.
Solution:
Let x be the boat’s speed, y be the current’s speed (in km. per min.)
x+y = 155 (1)
x−y = 205 (2) Subtracting (2) from (1): 2y= 13 −1
4 = 1 12
y = 12 121
A boat goes downstream a distance of 5 kilometers in 15 minutes but it takes 20 minutes for the return trip. Find the rate of the current.
Solution:
Let x be the boat’s speed, y be the current’s speed (in km. per min.)
x+y = 155 (1) x−y = 205 (2)
Subtracting (2) from (1): 2y= 13 −1 4 =
1 12
y = 12 121
A boat goes downstream a distance of 5 kilometers in 15 minutes but it takes 20 minutes for the return trip. Find the rate of the current.
Solution:
Let x be the boat’s speed, y be the current’s speed (in km. per min.)
x+y = 155 (1) x−y = 205 (2) Subtracting (2) from (1): 2y= 13 −1
4
= 121 y = 12 121
A boat goes downstream a distance of 5 kilometers in 15 minutes but it takes 20 minutes for the return trip. Find the rate of the current.
Solution:
Let x be the boat’s speed, y be the current’s speed (in km. per min.)
x+y = 155 (1) x−y = 205 (2) Subtracting (2) from (1): 2y= 13 −1
4 = 1 12
y = 12 121
A boat goes downstream a distance of 5 kilometers in 15 minutes but it takes 20 minutes for the return trip. Find the rate of the current.
Solution:
Let x be the boat’s speed, y be the current’s speed (in km. per min.)
x+y = 155 (1) x−y = 205 (2) Subtracting (2) from (1): 2y= 13 −1
4 = 1 12
y = 12 121
A boat goes downstream a distance of 5 kilometers in 15 minutes but it takes 20 minutes for the return trip. Find the rate of the current.
Solution:
Let x be the boat’s speed, y be the current’s speed (in km. per min.)
x+y = 155 (1) x−y = 205 (2) Subtracting (2) from (1): 2y= 13 −1
4 = 1 12
y = 12 121
1. With respect to which of the x-axis, y-axis, or origin is the graph ofy= xy
x−y symmetric? Justify your answer. 2. Find the equation of the line tangent to the circle
x2+y2−4x+ 6y+ 3 = 0 at the point (3,0).
3. Find the value/s of ksuch that the parabola with equation y=x2+kx+ 1 has no x-intercepts.
4. Find the solution set of the following: 4.1 |x2−5x−5|= 1
4.2 √√2x+ 3 =
x−2 +√x+ 1
4.3 2
x−1−
5x
2x2+x−3 ≤0 4.4
2x−3y+z =−1
−x+ 2y−3z =−2 7x−y−z =−8
4.5 (
x2+y2 = 5
x−y =−3
