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Write a quadratic equation with real number coefficients if one of the solutions in 3+4i?

+1 vote
There are more than 1 answer right?

the book got

f(x)= x^2 - 6x + 25
asked Feb 8, 2013 in ALGEBRA 1 by andrew Scholar

1 Answer

+1 vote

The real number coefficients is 3 + 4i then the other root has to be 3 - 4i because the coefficient of x is minus the sum of the roots and the independent term is the product of the roots and they would have imaginary parts.

So the equation is (x - 3 - 4i)(x - 3 + 4i)

= [(x - 3) - 4i ][(x - 3) + 4i]

Mathematics formula: (a + b)(a - b) = (a2 - b2), (a - b)2 = (a2 - 2ab + b2)

= (x - 3)2- (4i)2

Imaginary numbers in i2 = -1

= x2 - 2(x)(3) + 32 - (-16)

= x2 - 6x + 9 + 16

= x2 - 6x + 25.

answered Feb 8, 2013 by richardson Scholar

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