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.

tomtom Sep 7, 2023

#1**-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

ElectricPavlov Sep 7, 2023

#1**-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**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

--..-

Guest Sep 8, 2023

edited by
Guest
Sep 8, 2023