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Given that 2-3i is a root to the equation z^3+pz^2+qz-13p=0, show that p=-2 and q=5.

Substitute 2-3i into equation using part i (2-3i)3=-46-9i.  -46-9i+p(-5-12i)+q(2-3i)-13p=0. -46-18p+2q-9i-12pi-3iq=0. Real: -46-18p+2q=0 and Imaginary: -9-12p-3q=0. p=-2, q=5

WN
Answered by William N. Maths tutor
11313 Views

Show (2-3i)^3 can be expressed in the form a+bi where a and b are negative integers.

(2-3i) x (2-3i) = -5-12i.  -5-12i x (2-3i) = -46-9i.  a=-46, b=-9

WN
Answered by William N. Maths tutor
4293 Views

I'm struggling with approaching questions in Maths, I just don't know where to start. What should I do?

My approach will take you step by step through each type of question you find difficult. Maths (quite usefully) is very procedural. This means if you follow the instructions step by step you'll find you'r...

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Answered by Gus R. Maths tutor
4825 Views

State the nth term of the following sequence: 3, 7, 11, 15, 19

Start by labelling each term, e.g. 3 is the 1st, 7 is the second etc

Find the difference between each term, in this case +4 so we know our nth term will start with 4n 

Now we substitute in t...

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Answered by Alice L. Maths tutor
15507 Views

The number of questions I have to answer in the time is overwhelming! How can I make the most of my time during the UKCAT test?

The key to passing the UKCAT is to keep your momentum going throughout the test. Its easy to give up and trip into the pitfall of thinking you can't possibly do your best when your clock is slowly ticking...

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Answered by Gus R. UCAT tutor
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