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Quadratic are hard!

 

For what values of $j$ does the equation $(2x + 7)(x - 4) = -31 + jx$ have exactly one real solution? Express your answer as a list of numbers, separated by commas.

 Sep 7, 2023

Best Answer 

 #1
avatar+36745 
-2

2x^2 -x -28 = -31 + jx 

2x^2 - x - jx +3 = 0      for just one solution the discriminant of the quadrtatic equation must = 0 

b^2 - 4ac = 0 

(-j-1)^2 - 4 ( 2 )(3)  = 0 

     j^2 + 2j  - 23 = 0    Now use quadratic equation to find   j = -1 +2 sqrt6     or   ~~    3.89898    and - 5.89898  

 Sep 7, 2023
 #1
avatar+36745 
-2
Best Answer

2x^2 -x -28 = -31 + jx 

2x^2 - x - jx +3 = 0      for just one solution the discriminant of the quadrtatic equation must = 0 

b^2 - 4ac = 0 

(-j-1)^2 - 4 ( 2 )(3)  = 0 

     j^2 + 2j  - 23 = 0    Now use quadratic equation to find   j = -1 +2 sqrt6     or   ~~    3.89898    and - 5.89898  

ElectricPavlov Sep 7, 2023
 #2
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0

tomtom, you bloody idiot! You have asked plenty of questions already, and a great portion of them deal with quadratics and your ineptness at understanding them. One would expect that if you actually read the solutions and understood them that you would be able to deal with trivialities such as this. But no! You simply copy-paste the answer into the answer box and press 'submit' without giving the slightest flying damn about the time and effort you just wasted.

 

Granted, these homework simps are egotistical morons, but they are also somewhat competent mathematicians who could probably teach Year 3 math if given a degree...

 

GA

--..-

 Sep 8, 2023
edited by Guest  Sep 8, 2023
 #3
avatar+2437 
-6

For the record: Post #2, signed by GA, is not by the real GA.

 

GA

--. .- 

GingerAle  Sep 8, 2023
edited by GingerAle  Sep 8, 2023

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