Enable contrast version

# Tutor profile: Bilal H.

Inactive
Bilal H.
Student who has co-taught GCSE Further Maths for the past two years
Tutor Satisfaction Guarantee

## Questions

### Subject:Calculus

TutorMe
Question:

Find a general function for $$y$$ given, $$\frac{dy}{dx}$$$$=x\sqrt{y²-4}$$

Inactive
Bilal H.

Start by using a technique called "the separation of variables" where we isolate the $$x$$ and $$y$$ by multiplying both sides by $$dx$$ and dividing both sides by $$\sqrt{y²-4}$$ to give: $$\frac{1}{\sqrt{y²-4}}$$$$dy=xdx$$ We can then integrate both sides to remove the $$dy$$ and $$dx$$, giving: The integral of $$\frac{1}{\sqrt{y²-4}}$$$$dy=$$ the integral of $$xdx$$ these are both indefinite integrals. The LHS (integral with respect to $$y$$ is a bit more tricky and requires a hyperbolic substitution) by letting $$y=2cosh(u)$$ we find that $$dy=2sinh(u)du$$ this allows the integral to become $$\frac{1}{\sqrt{4cosh²(u)-4}}$$$$(2sinh(u))du$$ which simplifies to the integral of $$\frac{1}{2sinh(u)}$$$$2sinh(u)$$$$du$$ which is just the the integral of $$udu$$ which gives $$u+c$$ Integrating the RHS gives $$\frac{1}{2}$$$$x²+d$$ Thus finally yielding $$y+c=$$$$\frac{1}{2}$$$$x²+d$$ where $$c$$ and $$d$$ are arbirary constants. As both $$c$$ and $$d$$ are arbitrary constants this can be simplified to: $$y=$$$$\frac{1}{2}$$$$x²+A$$ where $$A$$ is an arbitrary constant.

### Subject:Physics

TutorMe
Question:

Explain, in molecular terms, why a pressure of a gas in a sealed container increases when the temperature of the container increases.

Inactive
Bilal H.

As the temperature of a gas increases, the mean speed of each particle also increases. This increases the number of collisions made by particles of the gas on the walls of the container per second. Thereby, the force on the walls of the container increases, and therefore the pressure increases.

### Subject:Algebra

TutorMe
Question:

Solve for all values of $$x$$ such that $$f(x)=0$$ $$x^3 + 5x^2 + 3x + 15$$

Inactive
Bilal H.

To begin, we want to find a factor of $$f(x)$$ in order to find a linear or quadratic term that we can factor out, allowing us to solve this more easily. Looking for terms in the form $$(x+a)$$ or $$(x²+bx+c)$$ Notice that in $$f(x)$$ the final (constant) term is 15. As it is easier I will start by assuming there exists some linear factor in the form $$(x+a)$$ that can be factored out of $$f(x)$$. In order to find this linear factor, I will use the factor theorem which states: if $$f(a)=0$$, therefore $$(x-a)$$ must be a factor of $$f(x)$$. The factors of 15 are: -15, -5, -3, -1, 1, 3, 5, 15. I find it easiest to work away from 0 and check $$f$$(each factor) to find where the output is 0. Check: $$f(1)=24$$ therefore $$(x-1)$$ is not a factor of $$f(x)$$ $$f(-1)=16$$ therefore $$(x+1)$$ is not a factor of $$f(x)$$ Eventually, I find that $$f(-5)=0$$ therefore $$(x+5)$$ must be a factor of $$f(x)$$ Next, I will divide $$f(x)$$ by $$(x+5)$$ in order to find a quadratic I can hopefully factor and then solve. $$\frac{f(x)}{(x+5)}$$=$$(x²+3)$$ which tells us that $$f(x) = (x+5)(x²+3)$$ As we are trying to find the values of $$x$$ such that $$f(x)=0$$ we can make $$(x+5)(x²+3)=0$$ This tells us that either $$(x+5)$$ or $$(x²+3)$$ are $$0$$ So we can consider them individually if $$(x+5)=0$$ therefore $$x=-5$$ If $$(x²+3)=0$$ then we get $$x²=-3$$ which is not solvable in the real world, however in the complex world we can solve this by taking the square root of both sides, leaving: $$x=$$$$±\sqrt{3}i$$ Therefore the answers to $$f(x)=0$$ are: $$x=-5$$, $$x=$$$$\sqrt{3}i$$ and $$x=$$$$-\sqrt{3}i$$

## Contact tutor

Send a message explaining your
needs and Bilal will reply soon.
Contact Bilal

## Request lesson

Ready now? Request a lesson.
Start Lesson

## FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage
Made in California by Zovio
© 2013 - 2021 TutorMe, LLC
High Contrast Mode
On
Off