Hi, I'm Josh!

I'm a fist year studying **C****hemical Engineering** at **M****anchester** 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 **C****hemical Engineering** at **M****anchester** 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!

No DBS Check

Suppose you have a quadratic equation: x^{2 }+ 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^{2 }+ 5x + 6

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

x^{2} + x**a** + x**b** + **ab** = x^{2 }+ 5x + 6

It's clear the x^{2} terms cancel out, and if we equate the x terms and the number terms, we are left with

x**a** + x**b** = 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) = x^{2} + 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^{2 }+ 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^{2 }+ 5x + 6

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

x^{2} + x**a** + x**b** + **ab** = x^{2 }+ 5x + 6

It's clear the x^{2} terms cancel out, and if we equate the x terms and the number terms, we are left with

x**a** + x**b** = 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) = x^{2} + 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