Видео

(x+6)^½-(2x-5)^½=2 (x+6)^½-2=(2x-5)^½ x+6-4(x+6)^½+4=2x-5 6-4(x+6)^½+4=x-5 10-4(x+6)^½=x-5 -4(x+6)^½=x-15 16(x+6)=x²-30x+225 16x+96=x²-30x+225 x²-46x+129=0 (x-3)(x-43)=0 x-3=0 x=3 ❤ (3+6)^½-[2(3)-5]^½=2 9^½-1^½=2 3-1=2 2=2 ☆ x-43=0 x=43 extraneous (43+6)^½-[2(43)-5]^½=2 49^½-81^½=2 7-9=2 -2≠2

Greetings. I got two answers, 43, 3. However, 43 yields a negative 2 whwn it is plugged into the original expression. Three is the only value that satisfies the expression.

Too slow n too long

You are good snr..!

The other one gave me the same answer...

J= ln(3/(e+1))

Thanks for making my understanding clear on completing square method . I used to struggle with it

For both sides, multiply by 3 --> square --> subtract 2.

What other approach can be use

I used to have problem when computing the values after integration. But now I how, thanks..

What all do you use?

will be waiting..

Impressive ❤

u need to be making this video 5 minutes long, not 50. People who want to go slower can slow the playback speed on RUclips. If u just did u = sqrt(x^2+9), then u^2 - 9 = x^2 and 2u du = 2x dx ==> u(u^2 - 9) du = x^3 dx, So your integral becomes over u^2 - 9 du which is really obvious. Could’ve saved yourself 50 minutes too

Are you fucking crazy to solve this at that rate.

If you take such a long time to solve this simple problem, you are bound to fail in all the exams....a tortoise will do this faster than you !

It is easier to substitute u² = x² + 9 which at once removes the square root and leads to a simple integration of (u² - 9)du

Good

What a stupid and long video.

Good.

Just by looking at it I can say x = 16

I like the approach used

Wow , u make the solving of indices so simple , keep up the good work

At first I used wonder why integral of sinx gives -cosx. But thanks to MIT maths I now know why. Waiting for more

Keep broadening our mind on calculus

cool we will be waiting

Thank God I now know how to use the indices method to solve questions of this nature. Waiting for more...