Question 176614
Find the quadratic equation whose roots are: {{{(4+i)}}} and {{{(4-i)}}}
Basically, we'll work in reverse of solving the quadratic equation, starting with:
{{{x = (4+i)}}} and {{{x = (4-i)}}} rewrite these as:
{{{x-(4+i) = 0}}} and {{{x-(4-i) = 0}}}...and we know that these are factors of the original quadratic equation, so we can reconstitute the equation by multiplying these factors. First, simplify the factors.
{{{(x-(4+i))*(x-(4-i)) = (x-4-i)*(x-4+i)}}} Now we'll perform the multiplication.
{{{(x-4-i)*(x-4+i) = x^2-4x+xi-4x+16-4i-xi+4i-i^2}}} Now we'll simplify this.
{{{x^2-8x+16-i^2 = x^2-8x+17}}}
The original quadratic equation is:
{{{highlight(x^2-8x+17 = 0)}}}
Let's check using the quadratic formula:{{{x = (-b+-sqrt(b^2-4ac))/2a}}}
{{{x = -(-8)+-sqrt((-8)^2-4(1)(17)))/2(1)}}}
{{{x = (8+-sqrt(64-68))/2}}}
{{{x = (8+-sqrt(-4))/2}}}
{{{x = (8/2)+2i/2}}} or {{{x = (8/2)-2i/2}}} Simplifying, we get:
{{{highlight(x = 4+i)}}} or {{{highlight(x = 4-i)}}}