Question -
Answer -
(i)┬аx2┬а+10ix тАУ 21 = 0
Given: x2┬а+10ix тАУ 21 = 0
x2┬а+10ix тАУ 21 ├Ч 1 = 0
We know, i2┬а=тАУ1┬атЗТ┬а1 = тАУi2
By substituting 1 = тАУi2┬аinthe above equation, we get
x2┬а+10ix тАУ 21(тАУi2) = 0
x2┬а+10ix + 21i2┬а= 0
x2┬а+3ix + 7ix + 21i2┬а= 0
x(x + 3i) + 7i(x + 3i)= 0
(x + 3i) (x + 7i) = 0
x + 3i = 0 or x + 7i =0
x = тАУ3i or тАУ7i
тИ┤ The roots of thegiven equation are тАУ3i, тАУ7i
(ii)┬аx2┬а+(1 тАУ 2i)x тАУ 2i = 0
Given: x2┬а+(1 тАУ 2i)x тАУ 2i = 0
x2┬а+ xтАУ 2ix тАУ 2i = 0
x(x + 1) тАУ 2i(x + 1) =0
(x + 1) (x тАУ 2i) = 0
x + 1 = 0 or x тАУ 2i =0
x = тАУ1 or 2i
тИ┤ The roots of thegiven equation are тАУ1, 2i
(iii)┬аx2┬атАУ(2тИЪ3 + 3i) x + 6тИЪ3i = 0
Given: x2┬атАУ(2тИЪ3 + 3i) x + 6тИЪ3i = 0
x2┬атАУ(2тИЪ3x + 3ix) + 6тИЪ3i = 0
x2┬атАУ2тИЪ3x тАУ 3ix + 6тИЪ3i = 0
x(x тАУ 2тИЪ3) тАУ 3i(x тАУ2тИЪ3) = 0
(x тАУ 2тИЪ3) (x тАУ 3i) = 0
(x тАУ 2тИЪ3) = 0 or (x тАУ3i) = 0
x = 2тИЪ3 or x = 3i
тИ┤ The roots of thegiven equation are 2тИЪ3, 3i
(iv)┬а6x2┬атАУ17ix тАУ 12 = 0
Given: 6x2┬атАУ17ix тАУ 12 = 0
6x2┬атАУ17ix тАУ 12 ├Ч 1 = 0
We know, i2┬а=тАУ1┬атЗТ┬а1 = тАУi2
By substituting 1 = тАУi2┬аinthe above equation, we get
6x2┬атАУ17ix тАУ 12(тАУi2) = 0
6x2┬атАУ17ix + 12i2┬а= 0
6x2┬атАУ9ix тАУ 8ix + 12i2┬а= 0
3x(2x тАУ 3i) тАУ 4i(2x тАУ3i) = 0
(2x тАУ 3i) (3x тАУ 4i) =0
2x тАУ 3i = 0 or 3x тАУ 4i= 0
2x = 3i or 3x = 4i
x = 3i/2 or x = 4i/3
тИ┤ The roots of thegiven equation are 3i/2, 4i/3