Question 422835
{{{8x^2+1 = 5x}}}
This is a quadratic equation so we want one side to be zero. Subtracting 5x from each side we get:
{{{8x^2-5x+1 = 0}}}
This won't factor so we'll use the Quadratic Formula:
{{{x = (-(-5) +- sqrt( (-5)^2 -4(8)(1)))/2(8)}}}
which simplifies as follows:
{{{x = (-(-5) +- sqrt( 25 -4(8)(1)))/2(8)}}}
{{{x = (-(-5) +- sqrt( 25 - 32))/2(8)}}}
{{{x = (-(-5) +- sqrt(-7))/2(8)}}}
{{{x = (5 +- sqrt(-7))/16}}}
With a negative number in the square root, we have complex solutions. To write our answers in standard form for complex numbers, a + bi, we start by factoring out -1:
{{{x = (5 +- sqrt(-1*7))/16}}}
{{{x = (5 +- sqrt(-1)*sqrt(7))/16}}}
Since {{{sqrt(-1) = i}}} this becomes:
{{{x = (5 +- i*sqrt(7))/16}}}
Now write the out the long way:
{{{x = (5 + i*sqrt(7))/16}}} or {{{x = (5 - i*sqrt(7))/16}}}
Now we split the fractions:
{{{x = 5/16 + (i*sqrt(7))/16}}} or {{{x = 5/16 + (-i*sqrt(7))/16}}}
which can be written as:
{{{x = 5/16 + (sqrt(7)/16)i}}} or {{{x = 5/16 + (-sqrt(7)/16)i}}}
These are the complex solutions written in standard form.