Joshua F. GCSE Chemistry tutor, GCSE Maths tutor, A Level Maths tutor...

Joshua F.

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Chemical Engineering w/ industrial experience (Masters) - Manchester University

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About me

Hi, I'm Josh!
I'm a fist year studying Chemical Engineering at Manchester currently and I sometimes get way too excited about mathematics... I enjoyed maths at school, so much so I taught myself an extra module (Mechanics 3) for which i got 90%. 

Besides acadaemic studies I'm a keen muscian, with piano, clarinet and voice up to grade 8, and I also like to keep myself busy playing guitar, saxophone and the organ. I'm a university rower and an avid cyclist, so I understand how difficult it can be to fit homework/coursewokrk around everything else in life! I've recently been awarded the BP STEM Scholarship which was given to only 90 students across the UK.

 

I hope I can make things easy to understand for anyone, and help you to achieve greatness! 

Hi, I'm Josh!
I'm a fist year studying Chemical Engineering at Manchester currently and I sometimes get way too excited about mathematics... I enjoyed maths at school, so much so I taught myself an extra module (Mechanics 3) for which i got 90%. 

Besides acadaemic studies I'm a keen muscian, with piano, clarinet and voice up to grade 8, and I also like to keep myself busy playing guitar, saxophone and the organ. I'm a university rower and an avid cyclist, so I understand how difficult it can be to fit homework/coursewokrk around everything else in life! I've recently been awarded the BP STEM Scholarship which was given to only 90 students across the UK.

 

I hope I can make things easy to understand for anyone, and help you to achieve greatness! 

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
PhysicsA-level (A2)A*
ChemistryA-level (A2)A
Further MathematicsA-level (A2)A

Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
PhysicsA Level£20 /hr
ChemistryGCSE£18 /hr
MathsGCSE£18 /hr
PhysicsGCSE£18 /hr
ScienceGCSE£18 /hr

Questions Joshua has answered

How to find the roots of a Quadratic Equation by Factorising?

Suppose you have a quadratic equation: x+ 5x + 6

You are told it has TWO REAL ROOTS and you are to find them.

The factorised quadratic with two roots will look like this:

(x+a)(x+b)

Where a and b are real numbers (any positive or negative number really)

As this is just another form of the original equation we can say they are equal:

(x+a)(x+b) = x+ 5x + 6

 

If you mulitply out the brackets using the FOIL (First, Outside, Inside, Last) rule you get:

x2 + xa + xb + ab = x+ 5x + 6

It's clear the x2 terms cancel out, and if we equate the x terms and the number terms, we are left with

xa + xb = 5x        and  ab= 6

x(a+b) = 5x                  a x b = 6

a + b = 5

So now we must use this information to find a and b.

The factors of 6 are:

6 and 1 

3 and 2

Of those factors, the pair which adds to 5 are 3 and 2.

so a = 3 and b = 2

Now we must check the signs of the factorised equation to check when we multiply ot the bracket we get the original equation again.... 

(x+3)(x+2) = x2 + 5x + 6 - CORRECT

Now we use this to find the roots, i.e. the x coordinates were y = 0 

so therefore we make our factorised quadratic equal to 0

(x+3)(x+2)=0

If two things multiplied together = 0, then at least one of them must equal 0...

x+3 = 0 ==> x = -3

OR

x+2 = 0 ==> x = -2

 

We can check the roots are corrects by replacing a and b with the x terms in the original equation and it should equal 0 for both a and b

We now have our roots...

x = -3 

and x = -2

Suppose you have a quadratic equation: x+ 5x + 6

You are told it has TWO REAL ROOTS and you are to find them.

The factorised quadratic with two roots will look like this:

(x+a)(x+b)

Where a and b are real numbers (any positive or negative number really)

As this is just another form of the original equation we can say they are equal:

(x+a)(x+b) = x+ 5x + 6

 

If you mulitply out the brackets using the FOIL (First, Outside, Inside, Last) rule you get:

x2 + xa + xb + ab = x+ 5x + 6

It's clear the x2 terms cancel out, and if we equate the x terms and the number terms, we are left with

xa + xb = 5x        and  ab= 6

x(a+b) = 5x                  a x b = 6

a + b = 5

So now we must use this information to find a and b.

The factors of 6 are:

6 and 1 

3 and 2

Of those factors, the pair which adds to 5 are 3 and 2.

so a = 3 and b = 2

Now we must check the signs of the factorised equation to check when we multiply ot the bracket we get the original equation again.... 

(x+3)(x+2) = x2 + 5x + 6 - CORRECT

Now we use this to find the roots, i.e. the x coordinates were y = 0 

so therefore we make our factorised quadratic equal to 0

(x+3)(x+2)=0

If two things multiplied together = 0, then at least one of them must equal 0...

x+3 = 0 ==> x = -3

OR

x+2 = 0 ==> x = -2

 

We can check the roots are corrects by replacing a and b with the x terms in the original equation and it should equal 0 for both a and b

We now have our roots...

x = -3 

and x = -2

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3 years ago

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