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Prove that an angle subtended by an arc is double at the centre then at the perimeter.

After drawing, bisect the perimeter angle in two from the centre. We now have two isosceles triangles with the sides the same length (1 radius from the centre). We know that 2 angles in each triangle add ...

Answered by Matthew P. • Maths tutor

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The complex conjugate of 2-3i is also a root of z^3+pz^2+qz-13p=0. Find a quadratic factor of z^3+pz^2+qz-13p=0 with real coefficients and thus find the real root of the equation.

z-2+3i times z-2-3i = z²-4z+13. z³-2z²+5z+26 divided by z²-4z+13 = z+2. Therefore the real root is z=-2.

Answered by William N. • Maths tutor

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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

Answered by William N. • Maths tutor

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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

Answered by William N. • Maths tutor

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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...

Answered by Gus R. • Maths tutor

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