Algebraically solve 3x-11√x-4=0?

Well first of all the way i am about to solve this will only work if the entire quantity (x-4) is under the square root.

Since you are multiplying 3x-11 by the square root, you have to divide to get rid of it.

Zero divided by 3x-11 equals 0, since zero divided by any number is zero.

Then you are left with √x-4=0

In order to get rid of the square root, you have to square the quantity under the square root.

So then you would have to square both sides and you would get x-4=0, since the square root and square would cancel and zero squared is zero.

Then add four to both sides and you get a final answer of x=4.

Hope this helps. It makes the most sense

the only that (she) used is named the subtitution approach... For me... that's an extremely puzzling to apply... i'm going to provide you an much less annoying approach.... the removing approach: as a manner to try this, coefficients (the selection in the previous the letter) of an identical variable could be same or opposite... if so: 3x + 7y = 5 6x + y = 5 observe which you will multiply the 1st equation to 2 to make: 6x + 14y = 10 6x + y = 5 considering the fact that there are same coefficients of x... (6x and 6x) you could subtract the two equations to get rid of x... 6x + 14y = 10 -(6x + y = 5) -------------------------- 13y = 5 y = 5/13 to unravel for x, you would be able to subtitute the cost of y (5/13) to the two of the two equations... in spite of the shown fact that it particularly is greater helpful to subtitute that to equation 2 (yet subtituting to eq. a million will additionally artwork) To equation 2: 6x + (5/13) = 5 Multiply the full equation to 13 to stay away from working with fractions... 78x + 5 = sixty 5 78x = sixty 5 - 5 78x = 60 x = 60/seventy 8 or 30/39 i'm going to attempt to equation a million.... 3x + 7(5/13) = 5 3x + 35/13 = 5 39x + 35 = sixty 5 39x = sixty 5-35 39x = 30 x = 30/39 (you will additionally get an identical) Pls mark me maximum suitable answer....