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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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Integrate by parts the following function: ln(x)/x^3

Let integrate be denoted by the letter I. For instance I(f) is the integration of a function f . Then Integration by parts states that I(u v') = uv - I(u' v), where u,v are function with u', v' their resp...

Answered by Paul D. • Maths tutor

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You are given the function f(x)=x^3-x^2-7x+3, and that x=3 is a root of f(x)=0. Find the exact values of the other 2 roots. (6 marks)

First step is to realise that as x=3 is a root of f(x)=0, then we can use (x-3) as a factor of f(x). A really good method to use to find what (fx)/(x-3) gives is Synthetic Division. Using this method we t...

Answered by David H. • Maths tutor

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How do you find the point or points of intersection of a straight line and a circle?

Given the equation of the straight line is: x+y=9 and the equation of the circle is: (x-2)²+(y-3)²=16 , find the point or points of intersection? The equation x+y=9 can be rearranged...

Answered by Edward C. • Maths tutor

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