i did the same🙂 nothing like solving a math problem to feel happy and satisfied🥰

Just a suggestion, I would love you try and solve a HSC Maths extention 2 paper in similar to the GSCE A level video you did previously.

For the no real solution suggestion, you can just add negative sign inside both log, i.e. log2(- 2x³ - 7x² - 2x - 3) = 3log2(- x - 1) + 1

@xinpingdonohoe3978

28 күн бұрын

I guess so. On the left you'd have +πki/ln(2) and on the right +3πki/ln(2) for some odd integer k (therefore non-zero), so it wouldn't be really feasible to try and equalise them.

I haven't even done log, and I aparantly didn't even need a log to solve this. It is way easier than it seems. When I saw you making both log 2, I thought yeah, the brackets are going to be the same. Thank you for explaining so well.

Even if there was some negative coefficients in that cubic polynomial, you could still trust that both solutions work in the real-number system as long as both values of x+1 are positive (because then 2(x+1)³ is also positive and that is equal to the complicated cubic polynomial for these values of x).

great video! pls solve some maths jee questions.

Agree one root could be negative and demand interpretation. It would be less obvious.

I'm totally disturbed by the way you solved that quadratic equation. I'm so used to X1,X2= (-b+-sqrt(b^2-4ac))/2a.

We still have to check the domain first to verify

you should check out CAPE pure mathematics unit 1 or unit 2

@saviplayer4546

5 күн бұрын

Aye I believe ik you

Sometimes, very elementary maths is required . Just draw an x,y table and join the dots. And presto There's your cubic😂😂😂😂😂

Wouldn't the negative answer still be a valid solution if the both sides get a complex result that are equal, and the negative answer is still real?

@hafizusamawrites

28 күн бұрын

Masha sense.

@oryxisatthefront8338

28 күн бұрын

How many REAL solutions x are there to the following equation?

@xinpingdonohoe3978

28 күн бұрын

​@@oryxisatthefront8338 yes. We want real x. But that doesn't impose on the original. For example, how many real solutions to √x=i are there? We find real x that satisfies. We aren't asked to find real x that satisfies and also keeps the original expression in R.

@Brid727

28 күн бұрын

the question never asked you to find all values of x, it only asked to find the number of REAL solutions that satisfy the equation once you find that, going beyond that is just a waste of your time so yeah but of course it would be a valid step if it were that the question asked to find all values of x

@xinpingdonohoe3978

28 күн бұрын

Yes. If they are real and solve it, that's good. Even if proving that they solve it means delving into the complex plane, they still solve it.

asnwer=1x isit hmm dhmm

if: ... =3 log(x - 1) + 1... NO real solution... 🤔

How can you be sure that the +1 at the end is base 2 and not base 10?

@bprpmathbasics

28 күн бұрын

Bc it says log_2

@joaomane4831

28 күн бұрын

Bro... Really?

@firstnamelastname4582

28 күн бұрын

You can basically do whatever you want with that last 1. 1 = log_2(2) = log_e(e) = log_10(10). But here the useful one is 1 = log_2(2) because the other log has base 2

@ronaldking1054

28 күн бұрын

It's arbitrary what 1 equals. There are many expressions. He just chose the one that helped him the best that did not violate any rules of equations. Things that would violate would be log with a base of a negative number of that negative number or square root of -1 times -1. Those are not in the domains of the functions that he is using.

this problem made me feel like I am good at math and I also can give this admission test. psss, nah! I suck at math, it was just the fact that this easy problem gave me overconfidence. Sorry, lol🤣🤣

Rewrite: Find the values of $a$ such that the following equation has two real solutions and only one real solution. $\log_2 \left( 2x^3+7x^2+2x+a ight) = 3 \log_2 \left( x + 1 ight) + 1$

@robertveith6383

22 күн бұрын

Don't bother to write Latex notation. It won't show up in special text in a KZread post, and, consequently, it's harder to reader because it is more messy.

I suppose that even if the solutions are negative, they still are real solutions. It requires the analytic continuation of logs to make sense of it, but the solutions would technically be real.