Question

if 4-i√3 is a root of quadratic equation then the equation is​

Answer 1

$\huge{\underline{\boxed{\green{\mathfrak{ANSWER:-}}}}}$

$= > x = 4 - i \sqrt{3} \\ = > x = 4 - \sqrt{3 \times {i}^{2} } \\ = > x - 4 = - \sqrt{ - 3} \: \: \: \: \: \: \: \: \: \: as \: {i}^{2} = - 1 \\ = > - ( x - 4) = \sqrt{ - 3} \\ = > { - ( x - 4)}^{2} = - 3 \\ {x}^{2} +16 -8x = + 3 \\ = > {x}^{2} - 8x + 13 = 0$

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Answer 2

Answer:

x2+8x+13=0

Step-by-step explanation:

4−i

3

is a root of the quadratic equation having real coefficients. Therefore, the other root of the equation is 4+i

3

. [Since, in a quadratic equation with real coefficients imaginary roots occur in conjugate(4−i

3

)+(4+i

3

)=8−i

3

)(4+i

3

)

=4

2

−(i

3

)

2

(∵x

2

−y

2

=(x+y)(x−y))

=16−3i

2

=16−(3×−1)(∵i 2 =−1)

=16+3

=19

product

2 − (sum of the roots)x+product of the roots =0x

2−8x+19=0

Hence, the equation is

x2 −8x+19=0.