write the quadratic equation having the following as roots: a.) 6-2i and 6+2i b.) 3 and 5

Asked on by tahiraha

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

a.) 6-2i and 6+2i

Since x1 and x2 are roots for the equation, then the factors are ( x-x1) and (x-x2)

==> f(x) = ( x - (6-2i) ( x - ( 6+2i)

==> f(x) = x^2 - (6+2i)x - (6-2i)x + ( 6-2i)(6+2i)

= x^2 - 6x - 2ix - 6x + 2ix + 36+ 4

= x^2 - 12x + 40

b.) 3 and 5

x1= 3     x2= 5

==> f(x) = (x-3) ( x-5)

= x^2 - 8x + 15

==> f(x) = x^2 - 8x + 15

neela | High School Teacher | (Level 3) Valedictorian

Posted on

There are two ways to construct a quadratic equation if  we are given the roots x1 and x2.

First method: (x-x1)(x-x2) = 0 and expand the left to get the equation whose roots are x1 and x2.

Second method : We assume  ax^2+bx+c = 0 is the equation and then by the relations between the roots and corfficients x1+x2 = -b/c and x1x2 = c/a, Now solve for b and c , choosing a = 1.

a) By first method.

x1 = 6-2i and x2 = 6+2i.

Therefore (x-1)(x-x2) = 0.

(x-(6-2i))(x+(x+(6+2i)) = 0.

{(x-6) +2i}{x-6-2i)} = 0.

(x-6)^2 - (2i)^2 = 0.

x^2 -12x+36 +4 = 0.

x^2 -12x +40 = 0 is the equation.

b) By second method:

Let ax^2+bx+c= 0 be the equation:

x1=3 and x2 = 5.

x1+x2 = -b/a and x1x2 = c/a.

3+5 = -b/a and  3*5 = c/a.

b= -8a and c= 15a.

Put a= 1. Then  b= -8a= -8, c = 15a = 15*1 = 15.

Therefore ax^2+bx+c = 0 becomes  1x^2-8x+15 = 0. Or

x^2-8x+15 = 0 is the equation which has the given roots 3 and 5.

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