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Urgent help with imaginary numbers?

Hi I'm taking further pure and i have a question about imaginary numbers.
It would be thankful if you'd helped me

Verify that 3+i is a root of the quadratic equation

z^2-(2+4i)z+8i-6=0

I'm not quite sure if z in the question is just a letter or z=x+iy

please help me

thanks
Reply 1
Original post by liemluji
Hi I'm taking further pure and i have a question about imaginary numbers.
It would be thankful if you'd helped me

Verify that 3+i is a root of the quadratic equation

z^2-(2+4i)z+8i-6=0

I'm not quite sure if z in the question is just a letter or z=x+iy

please help me

thanks


zz is a variable. It's like me telling you to check whether x=3x=3 is root of the quadratic equation x28x+15=0x^2 -8x + 15 = 0. You have two options of doing this, you can either substitute x=3x=3 into the LHS and see that it evaluates to 00 \,, or you could solve the quadratic equation and check that one of the roots is x=3x=3.

Use this logic here as well. Either plug z=3+iz =3+i into your quadratic and see if it evaluates to 00\, or solve it using the quadratic equation and see if one of your roots is indeed 3+i3+i.
(edited 8 years ago)
Reply 2
Original post by Zacken
zz is a variable. It's like me telling you to check whether x=3x=3 is root of the quadratic equation x28x+15=0x^2 -8x + 15 = 0. You have two options of doing this, you can either substitute x=3x=3 into the LHS and see that it evaluates to 00, or you could solve the quadratic equation and check that one of the roots is x=3x=3.

Use this logic here as well. Either plug z=3+iz =3+i into your quadratic and see if it evaluates to 00 or solve it using the quadratic equation and see if one of your roots is indeed 3+i3+i.


THank you so much! I managed to solve it
Reply 3
Original post by liemluji
THank you so much! I managed to solve it


Good work! Glad I helped.

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