So, there's no solution like that. Are you implying that it is faulty?. Just saying. I was thinking I can FOIL the right side and solve for x to clear the k first, but after getting the value of x, what am I supposed to do. I can solve for x. I know how to do it, but then I will need your help to go to the next step.Are you certain you copied the question correctly? The difficulty of the question is more challenging than the typical questions that you normally post.
See graphs here:https://www.desmos.com/calculator/8aetn7a3ob
Please double check, I think you didn't copy it correctly. There should be x2, not 62.Hi, dear tutors,
I find to find the value of k in this expression.
[math]6^2 + kx-5=(2x-1) (3x+5)[/math]
my question is,
Do I need to find the value of x first?.
thanks,
eddy
Oh, sorry. Yes, lev is right.Please double check, I think you didn't copy it correctly. There should be x2, not 62.
[math]x^2+kx-5=(2x-1)(3x+5)[/math]There are many solutions for k that satisfy the equation.Oh, sorry. Yes, lev is right.
[math]x^2[/math]
So, I'll take that to mean it is not a doable thing. Okay, thanks bbb[math]x^2+kx-5=(2x-1)(3x+5)[/math]There are many solutions for k that satisfy the equation.
See new graph: https://www.desmos.com/calculator/dkk03wah0m
It's doable, but k isn't just a singular value (without any constraint on x). What's the context of this problem?So, I'll take that to mean it is not a doable thing. Okay, thanks bbb
Just that. they ask to find the value ok in that expression. Nothing else.It's doable, but k isn't just a singular value (without any constraint on x). What's the context of this problem?
Please check or post the original problem. Is it possible the question is [math]\red{6x^2}+kx-5=(2x-1)(3x+5)[/math]Just that. they ask to find the value ok in that expression. Nothing else.
Yes, Bbb, very possible, there was a smudge at the beginning of the page. But confirming it with a friend, yes, it is that at the beginning, hence my initial 6.Please check or post the original problem. Is it possible the question is [math]\red{6x^2}+kx-5=(2x-1)(3x+5)[/math]
If so, the question is solvable for a singular value of k. Show your work of FOIL for the RHS.Yes, Bbb, very possible, there was a smudge at the beginning of the page. But confirming it with a friend, yes, it is that at the beginning, hence my initial 6.
[math]6x^2 + kx -5 = 6x^2 + 7x - 5[/math]Now, solve for k.Foiling the right-hand side.
[math](2x-1) (3x+5)[/math][math]expanding[/math][math](2x* 3x) + (2x* 5) + (-1 * 3x) + (-1 * 5)[/math] 6x^2 + 10x -3x -5 [math][/math]6x^2 + 7x -5[math]now,[/math] 6x^2 + kx -5 = 6x^2 + 7x - 5$$
This is the foiling of the right-hand side