How do you write a quadratic equation with #(-3 +- 2isqrt5)/4# as its roots, in the form of #ax^2 + bx +c = 0#, where #a, b, c# are integers?

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Answer 1 · 2016-04-05T09:24:31

Quadratic Equation
#16x^2+24x+29=0#

The given roots of the Quadratic Equation are

#x=-3/4+(2sqrt5i)/4# and #x=-3/4-(2sqrt5i)/4#

Transpose the complex numbers to the left side, so that right side is zero.

#x+3/4-(2sqrt5i)/4=0# and #x+3/4+(2sqrt5i)/4=0#

These are the factors Quadratic Equation which is equal to zero

#(x+3/4-(2sqrt5i)/4)(x+3/4+(2sqrt5i)/4)=0#

Just multiply and simplify this equation to obtain

#16x^2+24x+29=0#

God bless...I hope the explanation is useful.