Questions on Algebra: Quadratic Equation answered by real tutors!

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Question 150922: FInd the zeros of f(x)=x2 - 2X - 5: FInd the zeros of f(x)=x2 - 2X - 5
Answer by jim_thompson5910(8196) About Me  (Show Source):
You can put this solution on YOUR website!

x^2-2x-5=0 Start with the given equation.


Notice we have a quadratic equation in the form of ax^2+bx+c where a=1, b=-2, and c=-5


Let's use the quadratic formula to solve for x


x = (-b +- sqrt( b^2-4ac ))/(2a) Start with the quadratic formula


x = (-(-2) +- sqrt( (-2)^2-4(1)(-5) ))/(2(1)) Plug in a=1, b=-2, and c=-5


x = (2 +- sqrt( (-2)^2-4(1)(-5) ))/(2(1)) Negate -2 to get 2.


x = (2 +- sqrt( 4-4(1)(-5) ))/(2(1)) Square -2 to get 4.


x = (2 +- sqrt( 4--20 ))/(2(1)) Multiply 4(1)(-5) to get -20


x = (2 +- sqrt( 4+20 ))/(2(1)) Rewrite sqrt(4--20) as sqrt(4+20)


x = (2 +- sqrt( 24 ))/(2(1)) Add 4 to 20 to get 24


x = (2 +- sqrt( 24 ))/(2) Multiply 2 and 1 to get 2.


x = (2 +- 2*sqrt(6))/(2) Simplify the square root (note: If you need help with simplifying square roots, check out this solver)


x = (2)/(2) +- (2*sqrt(6))/(2) Break up the fraction.


x = 1 +- sqrt(6) Reduce.


x = 1+sqrt(6) or x = 1-sqrt(6) Break up the expression.


So our answers are x = 1+sqrt(6) or x = 1-sqrt(6)


which approximate to x=3.449 or x=-1.449