Given: 4x2 – 12x + 25 = 0
4x2 – 12x + 9 + 16 = 0
(2x)2 – 2(2x)(3) + 32 + 16 = 0
(2x – 3)2 + 16 = 0 [Since, (a + b)2 = a2 + 2ab + b2]
(2x – 3)2 + 16 × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
(2x – 3)2 + 16(–i2) = 0
(2x – 3)2 – 16i2 = 0
(2x – 3)2 – (4i)2 = 0
[By using the formula, a2 – b2 = (a + b) (a – b)]
(2x – 3 + 4i) (2x – 3 – 4i) = 0
2x – 3 + 4i = 0 or 2x – 3 – 4i = 0
2x = 3 – 4i or 2x = 3 + 4i
x = \( \dfrac{3}{2}\) – 2i or x = \( \dfrac{3}{2}\) + 2i
The roots of the given equation are \( \dfrac{3}{2}\) + 2i, \( \dfrac{3}{2}\) – 2i
Answered by Aaryan | 1 year agoShow that 1 + i10 + i20 + i30 is a real number?
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