Question 285242
{{{x^2=4x-29}}}
With quadratic equations you start by getting one side equal to zero. So we'll subtract 4x from each side and add 29 to each side:
{{{x^2-4x+29 = 0}}}
Then you factor it or use the Quadratic Formula. The expression on the left does not factor so we'll use the Quadratic Formula, with "a" = 1, "b" = -4 and "c" = 29:
{{{x = (-(-4) +- sqrt((-4)^2 - 4(1)(29)))/2(1)}}}
Simplifying we get:
{{{x = (4 +- sqrt(16 - 4(1)(29)))/2}}}
{{{x = (4 +- sqrt(16 - 116))/2}}}
{{{x = (4 +- sqrt(-100))/2}}}
Since the radicand (the expression within the square root) is negative we are going to end up with solutions that are complex numbers. If you haven't learned about complex numbers or if you are only interested in solutions that are real numbers, then we stop here and say that there are no real solutions to the equation.<br>
To find the complex solutions we factor out i (which is {{{sqrt(-1)}}}:
{{{x = (4 +- sqrt(100*(-1)))/2}}}
{{{x = (4 +- sqrt(100)*sqrt(-1))/2}}}
{{{x = (4 +- 10*i)/2}}}
Now we can split the plus or minus into two expressions:
{{{x = (4 + 10i)/2}}} or {{{x = (4 - 10i)/2}}}
Factoring out 2 in each numerator:
{{{x = (2(2 + 5i))/2}}} or {{{x = (2(2 - 5i))/2}}}
Canceling the factors of 2 we get:
{{{x = 2 + 5i}}} or {{{x = 2 - 5i}}}
which are the complex solutions to the equation.