Given: x2 – 4x + 7 = 0
x2 – 4x + 4 + 3 = 0
x2 – 2(x) (2) + 22 + 3 = 0
(x – 2)2 + 3 = 0 [Since, (a – b)2 = a2 – 2ab + b2]
(x – 2)2 + 3 × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
(x – 2)2 + 3(–i2) = 0
(x – 2)2 – 3i2 = 0
(x – 2)2 – \( (\sqrt{3i})^2\) = 0
[By using the formula, a2 – b2 = (a + b) (a – b)]
\( (x – 2 + \sqrt{3i}) (x – 2 – \sqrt{3i}) = 0\)
\( x = 2 ± \sqrt{3i}\)
The roots of the given equation are \( x = 2 ± \sqrt{3i}\)
Answered by Aaryan | 1 year agoShow that 1 + i10 + i20 + i30 is a real number?
Solve the quadratic equations by factorization method only 6x2 – 17ix – 12 = 0
Solve the quadratic equations by factorization method only x2 + (1 – 2i)x – 2i = 0
Solve the quadratic equations by factorization method only x2 + 10ix – 21 = 0
Solve the quadratic equations by factorization method only 17x2 – 8x + 1 = 0