find the value of k in the following algebraic expression

eddy2017

eddy2017

Elite Member
Joined
Oct 27, 2017
Messages
2,525
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
 
BigBeachBanana said:
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
Click to expand...
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.
 
eddy2017 said:
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
Click to expand...
Please double check, I think you didn't copy it correctly. There should be x2, not 62.
 
BigBeachBanana said:
Please check or post the original problem. Is it possible the question is [math]\red{6x^2}+kx-5=(2x-1)(3x+5)[/math]
Click to expand...
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.
 
eddy2017 said:
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.
Click to expand...
If so, the question is solvable for a singular value of k. Show your work of FOIL for the RHS.
 
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
 
eddy2017 said:
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
Click to expand...
[math]6x^2 + kx -5 = 6x^2 + 7x - 5[/math]Now, solve for k.
 
Let's solve for k.

[math]6x^2+kx−5=6x^2+7x−5[/math]
Addin’ -6x^2 to both sides.

[math]kx+6x^2−5+−6x^2= 6x^2+7x−5+−6x^2[/math]
[math]kx−5=7x−5[/math]
Addin’ 5 to both sides.

[math]kx−5+5=7x−5+5[/math]
[math]kx=7x[/math]
Dividin’ both sides by x.

[math]kx / x = 7x/x[/math]
[math]k=7[/I][/math]
 
You must log in or register to reply here.
Top