text: # Quadratic Equations Using a Formula
  
  • Jan 1st 2010, 10:53 AM
  ExponentialP
  Hi Happy New Year everyone hope you all have had a good one........ I'm doing Maths on New Years Day (Nerd)..... I'm having a bit of trouble with quadratic equations using a formula. I am using the formula to solve the equations but when I substitute the answers back in the equation the answers don't work....
  
  This is the equation :
  4²-3x-2=0
  
  I use the formula :
  x = -b ± √ b² - 4ac
  _____________
  2 a
  
  I substitute the letters for the numbers in the formula so that it looks like
  
  x = - - 3 ± √ -3² - 4 x 4 x -2
  _____________________
  2 x 4
  
  x = 3 + 4.80
  ________
  8
  
  or
  
  x = 3 - 4.80
  _________
  8
  
  Can someone please tell me where I'm going wrong
  
  I am stating roots correct to 2 dp
  • Jan 1st 2010, 11:00 AM
  Roam
  Your a=4, b=-3 and c=-2. So, $\sqrt{b^2 -4ac} = \sqrt{41} = 6.4$. The answer has to be $\frac{3 \pm \sqrt{41}}{8}$.
  • Jan 1st 2010, 11:02 AM
  earboth
  Quote:
  
  Originally Posted by ExponentialP
  Hi Happy New Year thank you
  everyone hope you all have had a good one........ I'm doing Maths on New Years Day (Nerd)..... I'm having a bit of trouble with quadratic equations using a formula. I am using the formula to solve the equations but when I substitute the answers back in the equation the answers don't work....
  
  This is the equation :
  4²-3x-2=0
  
  I use the formula :
  x = -b ± √ b² - 4ac
  _____________
  2 a
  
  I substitute the letters for the numbers in the formula so that it looks like
  
  x = - - 3 ± √( -3)² - 4 x 4 x -2
  _____________________
  2 x 4
  
  x = 3 + 4.80
  ________
  8
  
  or
  
  x = 3 - 4.80
  _________
  8
  
  Can someone please tell me where I'm going wrong
  
  I am stating roots correct to 2 dp
  
  See corrections.
  • Jan 1st 2010, 11:03 AM
  ExponentialP
  Yes thats what I done, but my answers still don't add up
  • Jan 1st 2010, 11:09 AM
  earboth
  Quote:
  
  Originally Posted by ExponentialP
  Yes thats what I done, but my answers still don't add up
  
  If you calculate
  
  $\sqrt{(-3)^2-4 \cdot 4 \cdot (-2)} = \sqrt{41}\approx 6.403$
  
  but you still use $\sqrt{23} \approx 4.8$
  • Jan 1st 2010, 11:11 AM
  ExponentialP
  Thank u earbooth am I doing something wrong at this point
  
  x = 3 +6.40
  ________
  8
  
  or
  
  x = 3 - 6.40
  _________
  8
  • Jan 1st 2010, 11:26 AM
  e^(i*pi)
  Quote:
  
  Originally Posted by ExponentialP
  Thank u earbooth am I doing something wrong at this point
  
  x = 3 +6.40
  ________
  8
  
  or
  
  x = 3 - 6.40
  _________
  8
  
  Both are equally valid solutions. Sub them back into the original equation if you want to be sure (Wink)
  • Jan 2nd 2010, 05:27 AM
  ExponentialP
  I think I might be doing something wrong somewhere, I seem to be getting answers that don't make sense when I apply the formula to the equation, for instance take :
  
  2x² - 7x = 3
  
  The exercise I am doing is asking me to solve this equation using a formula... I'm not sure if this is where I am going wrong but I use
  
  a= 2 b= -7 c= 0 ????
  
  I then get
  
  x = - (-7)± √ -7² - 4x2x0
  ____________________
  2 x 2
  
  The answer I get from this does not make sense, x = 7 ± √ -49
  ___________
  4
  
  Thanks
  
  Happy New Year....... Again
  • Jan 2nd 2010, 05:41 AM
  earboth
  Quote:
  
  Originally Posted by ExponentialP
  I think I might be doing something wrong somewhere, I seem to be getting answers that don't make sense when I apply the formula to the equation, for instance take :
  
  2x² - 7x = 3
  
  The exercise I am doing is asking me to solve this equation using a formula... I'm not sure if this is where I am going wrong but I use
  
  a= 2 b= -7 c= 0 ????
  
  I then get
  
  x = - (-7)± √ -7² - 4x2x0
  ____________________
  2 x 2
  
  The answer I get from this does not make sense, x = 7 ± √ -49
  ___________
  4
  
  Thanks
  
  Happy New Year....... Again
  
  If you want to apply the quadratic formula the equation must be in the form
  
  $ax^2+bx+c=0$
  
  Therefore you have to transform the equation first:
  
  $2x^2 - 7x = 3~\implies~\boxed{2x^2-7x-3=0}$
  
  So you have a = 2, b = -7, c = -3.
  • Jan 2nd 2010, 05:45 AM
  ExponentialP
  Thank u !!!
  • Jan 2nd 2010, 06:03 AM
  ExponentialP
  But I can't square root a negative number ??? What happens then ???
  • Jan 2nd 2010, 06:30 AM
  skeeter
  Quote:
  
  Originally Posted by ExponentialP
  But I can't square root a negative number ??? What happens then ???
  
  $2x^2 - 7x - 3 = 0$
  
  $b^2 - 4ac = (-7)^2 - 4(2)(-3) = 49+24 = 73$
  
  $x = \frac{7 \pm \sqrt{73}}{4}$
explanation: The text extract contains a math question and a series of responses that provide an answer and further clarification. The initial question is about solving a quadratic equation using the quadratic formula, and the user encounters difficulties with the calculations. Other users provide corrections and explanations to help the user understand where they went wrong.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: # Quadratic Equations Using a Formula
  
  • Jan 1st 2010, 10:53 AM
  ExponentialP
  Hi Happy New Year everyone hope you all have had a good one........ I'm doing Maths on New Years Day (Nerd)..... I'm having a bit of trouble with quadratic equations using a formula. I am using the formula to solve the equations but when I substitute the answers back in the equation the answers don't work....
  
  This is the equation :
  4²-3x-2=0
  
  I use the formula :
  x = -b ± √ b² - 4ac
  _____________
  2 a
  
  I substitute the letters for the numbers in the formula so that it looks like
  
  x = - - 3 ± √ -3² - 4 x 4 x -2
  _____________________
  2 x 4
  
  x = 3 + 4.80
  ________
  8
  
  or
  
  x = 3 - 4.80
  _________
  8
  
  Can someone please tell me where I'm going wrong
  
  I am stating roots correct to 2 dp
  • Jan 1st 2010, 11:00 AM
  Roam
  Your a=4, b=-3 and c=-2. So, $\sqrt{b^2 -4ac} = \sqrt{41} = 6.4$. The answer has to be $\frac{3 \pm \sqrt{41}}{8}$.
  • Jan 1st 2010, 11:02 AM
  earboth
  Quote:
  
  Originally Posted by ExponentialP
  Hi Happy New Year thank you
  everyone hope you all have had a good one........ I'm doing Maths on New Years Day (Nerd)..... I'm having a bit of trouble with quadratic equations using a formula. I am using the formula to solve the equations but when I substitute the answers back in the equation the answers don't work....
  
  This is the equation :
  4²-3x-2=0
  
  I use the formula :
  x = -b ± √ b² - 4ac
  _____________
  2 a
  
  I substitute the letters for the numbers in the formula so that it looks like
  
  x = - - 3 ± √( -3)² - 4 x 4 x -2
  _____________________
  2 x 4
  
  x = 3 + 4.80
  ________
  8
  
  or
  
  x = 3 - 4.80
  _________
  8
  
  Can someone please tell me where I'm going wrong
  
  I am stating roots correct to 2 dp
  
  See corrections.
  • Jan 1st 2010, 11:03 AM
  ExponentialP
  Yes thats what I done, but my answers still don't add up
  • Jan 1st 2010, 11:09 AM
  earboth
  Quote:
  
  Originally Posted by ExponentialP
  Yes thats what I done, but my answers still don't add up
  
  If you calculate
  
  $\sqrt{(-3)^2-4 \cdot 4 \cdot (-2)} = \sqrt{41}\approx 6.403$
  
  but you still use $\sqrt{23} \approx 4.8$
  • Jan 1st 2010, 11:11 AM
  ExponentialP
  Thank u earbooth am I doing something wrong at this point
  
  x = 3 +6.40
  ________
  8
  
  or
  
  x = 3 - 6.40
  _________
  8
  • Jan 1st 2010, 11:26 AM
  e^(i*pi)
  Quote:
  
  Originally Posted by ExponentialP
  Thank u earbooth am I doing something wrong at this point
  
  x = 3 +6.40
  ________
  8
  
  or
  
  x = 3 - 6.40
  _________
  8
  
  Both are equally valid solutions. Sub them back into the original equation if you want to be sure (Wink)
  • Jan 2nd 2010, 05:27 AM
  ExponentialP
  I think I might be doing something wrong somewhere, I seem to be getting answers that don't make sense when I apply the formula to the equation, for instance take :
  
  2x² - 7x = 3
  
  The exercise I am doing is asking me to solve this equation using a formula... I'm not sure if this is where I am going wrong but I use
  
  a= 2 b= -7 c= 0 ????
  
  I then get
  
  x = - (-7)± √ -7² - 4x2x0
  ____________________
  2 x 2
  
  The answer I get from this does not make sense, x = 7 ± √ -49
  ___________
  4
  
  Thanks
  
  Happy New Year....... Again
  • Jan 2nd 2010, 05:41 AM
  earboth
  Quote:
  
  Originally Posted by ExponentialP
  I think I might be doing something wrong somewhere, I seem to be getting answers that don't make sense when I apply the formula to the equation, for instance take :
  
  2x² - 7x = 3
  
  The exercise I am doing is asking me to solve this equation using a formula... I'm not sure if this is where I am going wrong but I use
  
  a= 2 b= -7 c= 0 ????
  
  I then get
  
  x = - (-7)± √ -7² - 4x2x0
  ____________________
  2 x 2
  
  The answer I get from this does not make sense, x = 7 ± √ -49
  ___________
  4
  
  Thanks
  
  Happy New Year....... Again
  
  If you want to apply the quadratic formula the equation must be in the form
  
  $ax^2+bx+c=0$
  
  Therefore you have to transform the equation first:
  
  $2x^2 - 7x = 3~\implies~\boxed{2x^2-7x-3=0}$
  
  So you have a = 2, b = -7, c = -3.
  • Jan 2nd 2010, 05:45 AM
  ExponentialP
  Thank u !!!
  • Jan 2nd 2010, 06:03 AM
  ExponentialP
  But I can't square root a negative number ??? What happens then ???
  • Jan 2nd 2010, 06:30 AM
  skeeter
  Quote:
  
  Originally Posted by ExponentialP
  But I can't square root a negative number ??? What happens then ???
  
  $2x^2 - 7x - 3 = 0$
  
  $b^2 - 4ac = (-7)^2 - 4(2)(-3) = 49+24 = 73$
  
  $x = \frac{7 \pm \sqrt{73}}{4}$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract contains a math question and a series of responses that provide an answer and further clarification. The initial question is about solving a quadratic equation using the quadratic formula, and the user encounters difficulties with the calculations. Other users provide corrections and explanations to help the user understand where they went wrong.
  
  Label: question-and-answer