2 questions (1 Viewer)

[ronaldinho]

x^2 - 4x + 5 (answer says 2,2 what does this mean?)

ix^2 - x + 4i = 0 answer is : 1/2(1+rt17)i



+ or - rt 17

 

[Slidey]

For the first question: that means nothing. The roots are complex and the vertex is (2,1).

For the second: I got a complex root, and it was entirely imaginary, but I didn't get sqrt17.

 

[bos1234]

For the first question: that means nothing. The roots are complex and the vertex is (2,1).

For the second: I got a complex root, and it was entirely imaginary, but I didn't get sqrt17.

ix^2 - x + 4i = 0 answer is : 1/2(1+rt17)i

1 + rt (1-4i.4i) divide by 2i

1 + rt17 divide by 2i

i keep getting this..

 

[Slidey]

Well the answers got that, too, so I would wager a guess that you're right. Pardon my drunken mental calculations and well done.

 

[Riviet]

ix^2 - x + 4i = 0 answer is : 1/2(1+rt17)i

1 + rt (1-4i.4i) divide by 2i

1 + rt17 divide by 2i

i keep getting this..

Looks right to me. So I'd say the answers are wrong.

(2,2) => the point on the argand diagram representing 2 + 2i, but it's wrong. Answer should be (2,1) as Slidey has pointed out.

 

[ronaldinho]

If a,b,c are real and b^2 < 4ac, show that the roots of ax^2 + bx + c = 0 are complex conjugates

 

[pLuvia]

If a,b,c are real and b^2 < 4ac, show that the roots of ax^2 + bx + c = 0 are complex conjugates

Just use the discriminant
disc.=b²-4ac
And since b²<4ac, hence the discriminant is less than zero and so the roots of that equation are complex roots. And complex roots appear in conjugates

 

[ronaldinho]

 

[jyu]

:wave:

 

[bos1234]

nice... real nice especially the 2nd qn working out!

ey what did u put in ur first line in the first qn.. its hard to c

let i^n = i^?? where k=012 m = 012...

 

[jyu]

:wave: