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barxatty [35]
2 years ago
11

What are the exact solutions of x2 - 5x - 1 = 0?.

Mathematics
2 answers:
borishaifa [10]2 years ago
8 0
x^2-5x-1=0\\\\a=1;\ b=-5;\ c=-1\\\Delta=b^2-4ac\\\\\Delta=(-5)^2-4\cdot1\cdot(-1)=25+4=29 \ \textgreater \  0\\\\therefore\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ \dfrac{-b+\sqrt\Delta}{2a}\\\\x_1=\dfrac{5-\sqrt{29}}{2\cdot1}=\boxed{\dfrac{5-\sqrt{29}}{2}}\\\\x_2=\dfrac{5+\sqrt{29}}{2\cdot1}=\boxed{\dfrac{5+\sqrt{29}}{2}}
masya89 [10]2 years ago
5 0

Answer:

Please use math-way

the person below or above is wrong

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( – 1, – 5) 7x+2y=13, 4x+4y=14
serious [3.7K]

Answer:

{x,y} = {6/5,23/10}

Step-by-step explanation:

 [1]    7x + 2y = 13

  [2]    4x + 4y = 14  <---------- linear equations given

Graphic Representation of the Equations : PICTURE
2y + 7x = 13        4y + 4x = 14  

Solve by Substitution :

// Solve equation [2] for the variable  y

[2]    4y = -4x + 14

[2]    y = -x + 7/2

// Plug this in for variable  y  in equation [1]

  [1]    7x + 2•(-x +7/2) = 13

  [1]    5x = 6

// Solve equation [1] for the variable  x

  [1]    5x = 6

  [1]    x = 6/5

// By now we know this much :

   x = 6/5

   y = -x+7/2

// Use the  x  value to solve for  y

   y = -(6/5)+7/2 = 23/10

// Plug this in for variable  y  in equation [1]

 [1]    7x + 2•(-x +7/2) = 13

  [1]    5x = 6

// Solve equation [1] for the variable  x

 [1]    5x = 6

[1]    x = 6/5

// By now we know this much :

   x = 6/5

   y = -x+7/2

// Use the  x  value to solve for  y

  y = -(6/5)+7/2 = 23/10

4 0
2 years ago
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