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Use the quadratic formula to solve each of the following equations.

0 votes
  1. x2+2x=7
  2. x2+2x=7
  3. 3q2+11=5q
  4. 3q2+11=5q
  5. 7t2=6−19t
asked Oct 16, 2018 in ALGEBRA 2 by anonymous

1 Answer

0 votes
1)
x^2 + 2x  =  7
x^2 + 2x  - 7  =  0
Compare it to quadratic form ax^2 + bx + c  =  0
a = 1, b = 2, c = -7
Roots are  
x  =  [ - b ± √(b^2 - 4ac) ] / 2a
x  =  [ - (2) ± √((2)^2 - 4(1)(-7)) ] / 2(1)
x  =  [ - 2 ± √(4 + 28) ] / 2
x  =  [ - 2 ± √32 ] / 2
x  =  [ - 2 ± 4√2 ] / 2
x  =  [ - 2 + 4√2 ] / 2       ;        x  =  [ - 2 - 4√2 ] / 2
x  =  - 2/2 + (4√2)/2       ;        x  =  - 2/2 - (4√2)/2
x  =  - 1 + 2√2       ;        x  =  - 1 - 2√2
2)
3q^2 + 11  =  5q
3q^2 + 11 - 5q  =  0
3q^2 - 5q + 11  =  0
Compare it to quadratic form ax^2 + bx + c  =  0
a = 3, b = -5, c = 11
Roots are  
x  =  [ - b ± √(b^2 - 4ac) ] / 2a
x  =  [ - (-5) ± √((-5)^2 - 4(3)(11)) ] / 2(3)
x  =  [ 5 ± √(25 - 132) ] / 6
x  =  [ 5 ± √(-107) ] / 6
x  =  [ 5 ± i√107 ] / 6
x  =  [ 5 + i√107 ] / 6      ;     x  =  [ 5 - i√107 ] / 6
x  =   5/6 + i(√107)/6      ;     x  =  5/6 - i(√107)/6
3)
7t^2  =  6 - 19t
​7t^2 - 6 + 19t  =  0
​7t^2 + 19t - 6  =  0
Compare it to quadratic form ax^2 + bx + c  =  0
a = 7, b = 19, c = -6
Roots are  
x  =  [ - b ± √(b^2 - 4ac) ] / 2a
x  =  [ - (19) ± √((19)^2 - 4(7)(-6)) ] / 2(7)
x  =  [ - 19 ± √(361 + 168) ] / 14

x  =  [ - 19 ± √529 ] / 14

x  =  [ - 19 ± 23 ] / 14
x  =  [ - 19 + 23 ] / 14      ;          x  =  [ - 19 - 23 ] / 14
x  =  4 / 14                      ;          x  =  -42 / 14
x  =  2 / 7                        ;          x  = - 3
Answer :

1)  The Solutions are x  =  - 1 + 2√2  and   x  =  - 1 - 2√2

2)  The Solutions are  x  =   5/6 + i(√107)/6  and  x  =  5/6 - i(√107)/6

3)  The Solutions are  x  =  2/7  and  x =- 3.

 

answered Oct 18, 2018 by homeworkhelp Mentor
edited Oct 21, 2018 by bradely

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