Question

If one of the values of x of the equation 2x^2-6x+k=0 be 1/2(a+5i), find the values of a & k.

Answer 1

is that
\[\huge \frac12(a+5i)\]?

Answer 2

ya

Answer 3

Okay, you should know that whenever a complex number is a root of some polynomial with real coefficients, then its conjugate would also be a root. So
\[\huge \frac12(a-5i)\] is also a root.

Answer 4

ya i know that

Answer 5

So... You will kindly evaluate this for me...
\[\huge \left[x-\frac12(a-5i)\right]\left[x-\frac12(a+5i)\right]\]
Take your time :)

Answer 6

x^2-xa+1/4a^2+25/4

Answer 7

Not bad :D
So, the coefficient of x^2 in your polynomial is 2, right? So you better multiply everything by 2... what do you get?

Answer 8

\[\huge 2x^2-2ax+\frac12a^2+\frac{25}2\]

Answer 9

ya i got that but tell me the final approach

Answer 10

I'm getting there :)
\[\Large \color{green}2x^2\color{red}{-2a}x+\color{blue}{\frac12a^2+\frac{25}2}=\color{green}2x^2\color{red}{ - 6}x + \color{blue}k\]

Answer 11

ya got this too.... but tell me the approach.....i mean why did we multiply [x−1/2(a−5i)][x−1/2(a+5i)] in the first place what was the approach

Answer 12

The logic is we're working on the equation that would have both
\[\large \frac12(a+5i)\] and \[\large \frac12(a-5i)\] as roots.

Now, when you have a quadratic equation with roots at a and b, you can bet that that quadratic function can be factored in the form
\[\huge p(x-a)(x-b)\]
where p is a constant.

Answer 13

Can you do it from here?

Answer 14

ya its done already......

Answer 15

thanks a lot.....terence

Answer 16

So, curiosity strikes... what values for a and k?

Answer 17

a=3

Answer 18

k=17 as 25/2= 12.5 and 1/2 * 9 = 4.5 and 12.5+4.5 is 17