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From users who are members of Transition to university :
Christian Lawson-Perfect
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said | Ready to use | 7 years, 4 months ago |
| Hannah Aldous | said | Needs to be tested | 7 years, 4 months ago |
Chris Graham
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said | Has some problems | 7 years, 4 months ago |
| Vicky Hall | said | Has some problems | 7 years, 4 months ago |
History
Elliott Fletcher 7 years, 4 months ago
Published this.
Christian Lawson-Perfect 7 years, 4 months ago
Gave some feedback: Ready to use
Hannah Aldous 7 years, 4 months ago
Gave some feedback: Needs to be tested
Christian Lawson-Perfect 7 years, 4 months ago
Gave some feedback: Has some problems
Christian Lawson-Perfect 7 years, 4 months ago
Saved a checkpoint:
The marking is set up wrong: $k(-6)$ is the application of the function $k$, not $-6 \times k$. You can avoid this by writing $(\ldots)k$ instead of $k(\ldots)$.
It's not obvious what order you should enter the roots. The best I can come up with for a hint is "Assuming $k$ is positive, enter the lowest root first." I can't think how to be any clearer without giving a hint about the form of the answer.
You need \left and \right around brackets to make sure they stretch to fit around fractions.
Put the quadratic formula in the advice just before you list the values for $a$, $b$ and $c$.
Chris Graham 7 years, 4 months ago
Gave some feedback: Has some problems
Chris Graham commented 7 years, 4 months ago
When the variable C_2 is zero, your advice states $c = 0k^2$, which caused me great confusion (about where the $k^2$ had come from) when I was unlucky enough to generate this on my first attempt. If you would like to keep this scenario, then you will need to adjust the advice, or if it is not desirable then use an
exceptin the generation of B_2 or n2.
Hannah Aldous 7 years, 4 months ago
Gave some feedback: Needs to be tested
Vicky Hall 7 years, 4 months ago
Gave some feedback: Has some problems
Vicky Hall commented 7 years, 4 months ago
This is a well conceived question but the answers look a bit nasty! It would be a much nicer question if, when we get to the point of having $\sqrt(xk^2)$, the $x$ is a square number. That way we end up with some quantity of $k$ divided by $2$, which may or may not simplify, but it looks a lot better than having a surd or a decimal.
Hannah Aldous 7 years, 4 months ago
Gave some feedback: Needs to be tested
Hannah Aldous 7 years, 4 months ago
Created this as a copy of Using the Quadratic Formula to Solve Equations of the Form $ax^2 +bx+c=0$.There are 58 other versions that do you not have access to.
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