Transition to university

Vicky Hall commented on Using the Quadratic Formula to Solve Equations of the Form $ax^2 +bx+c=0$ 8 years, 5 months ago

I would give the quadractic formula in the statement and have $x=$ in front of it. I would amend the statement so that it says it can also be useful to use the formula if equations are difficult to factorise (perhaps if coefficients are large), as the equations in part a) and part b)i) could both be solved by factorising instead but using the formula is (probably) quicker. 

The expected answers in part a) are the numbers that would appear in the factorised equation and not the roots so you need to negate the answers and then swap their gaps around.

Bradley Bush commented on Extract common factors of polynomials 8 years, 5 months ago

Thank you for the feedback, I should have fixed the marking now and improed the advice. The brackets do make everyhing look clearer. I have also included all the *'s especially seeing as they're needed for the question to be marked right now.

I havent put full stops at the end of EVERY calculation because it just feel unnecissary where they aren't part of a statement.

Stanislav Duris commented on Calculate the original price before a decrease 8 years, 5 months ago

Thank you Hannah. I removed the first part of my advice.

Stanislav Duris on Calculate the original price before a decrease 8 years, 5 months ago

Gave some feedback: Needs to be tested

Vicky Hall on Decimals to fractions 8 years, 5 months ago

Gave some feedback: Has some problems

Vicky Hall commented on Decimals to fractions 8 years, 5 months ago

I think this should be called 'converting decimals to fractions'. Do you have or are you going to make another question on converting fractions to decimals?

The numbers are very large in part b) of the questions. My first fraction was $\displaystyle\frac{681}{10000}$ and my second was $\displaystyle\frac{4365}{999}$. I can see fairly easily that $\displaystyle\frac{681}{10000}$ has no common factor but I had to check with my calculator if $3$ was a factor of $4365$. To fix this, I would reduce the decimal places, so perhaps make the non-recurring decimal $2$ decimal places long, and have the recurring decimal repeat after either $1$ or $2$ decimal places.

Make sure all numbers in the advice are in Latex. In the working for the recurring decimals, it would look much neater to have the $=$ aligned in the subtraction.

I have fixed the notation issue for the recurring decimal in part a).

Stanislav Duris on Rounding and estimating calculations 2 8 years, 5 months ago

Gave some feedback: Needs to be tested

Hannah Aldous on Factorising Quadratic Equations with $x^2$ Coefficients of 1 8 years, 5 months ago

Gave some feedback: Needs to be tested

Vicky Hall commented on Find and use the formula for a geometric sequence 8 years, 5 months ago

It would also be helpful to give and exlain the formula for the $n$th term in the statement.

Vicky Hall on Find and use the formula for a geometric sequence 8 years, 5 months ago

Gave some feedback: Has some problems

Vicky Hall commented on Find and use the formula for a geometric sequence 8 years, 5 months ago

I have added a sentence to the statement to explain what is meant by the common ratio. I think part a) should include one sequence with a positive integer ratio, one with a negative integer ratio and one with a fractional ratio to cover all possibilities.

Elliott Fletcher on Finding the formula for the $n^{\text{th}}$ term of linear sequences 8 years, 5 months ago

Gave some feedback: Has some problems

Elliott Fletcher commented on Finding the formula for the $n^{\text{th}}$ term of linear sequences 8 years, 5 months ago

Main Parts

I think you should label each of the questions i) and ii) as you have done in the advice section

I think the expected answer in both of these questions is wrong, the answer that you have in the advice section is right, i think its just that you calculated the adding term using n=0 for the displayed answer.

Advice

i) where you have written nth term here it doesn't really display correctly, it looks like n^(t)h instead of n^th.

I think you should say "Our nth term will take the form 8n+b, where b is some constant"

ii) i think "them" should be "it"

I think you should say "Our nth term will take the form 5n+c, where c is some constant"

I think you use the wrong number when you say how much the original sequence is greater than the sequence for 5n by. 

Vicky Hall on Finding the formula for the $n^{\text{th}}$ term of linear sequences 8 years, 5 months ago

Gave some feedback: Has some problems

Vicky Hall commented on Finding the formula for the $n^{\text{th}}$ term of linear sequences 8 years, 5 months ago

Change the question to 'find the formula for the $n$th term'. It would be good to have one increasing and one decrasing sequence to show that sometimes the $n$ coefficient is negative.

Vicky Hall on Finding the $n^{\text{th}}$ Term of a Quadratic Sequence 8 years, 5 months ago

Gave some feedback: Has some problems

Vicky Hall commented on Finding the $n^{\text{th}}$ Term of a Quadratic Sequence 8 years, 5 months ago

I would reword the question to say 'Find the formula for the $n$th term of the quadratic sequence'.

The advice is good but think the last part, where you find the constant, would look better if it were laid out like the previous part, so you show the student what the sequence $3n^2+3n$ looks like compared to the original sequence. You also need to remove the reference to part b).

Vicky Hall on Factorising Quadratic Equations with $x^2$ Coefficients Greater than 1 8 years, 5 months ago

Gave some feedback: Has some problems