SOLUTION: how do you do this (-b±√(b^2-4ac)2a
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Question 425871: how do you do this (-b±√(b^2-4ac)2a
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
sandard form of a quadratic equation is:
ax^2 + bx + c = 0
a is the coefficient of the x^2 term
b is the coefficient of the x term
c is the constant term
a,b,c are use in he equation x = (-b±√(b^2-4ac)2a
you replace a,b,c in the equation with the value you derived from the standard form of your equation.
that provides you with the x value of when the graph of the equation crosses the x axis.
an example
your equation is x^2 + 3x - 10 = 0
this equation is already in standard form of ax^2 + bx + c = 0
this gives you:
a = 1
b = 3
c = -10
plug those values into your quadratic equation and you get:
x =
simplify this equation to get:
x =
solve for x to get:
x = -5 or x = +2
those are the roots of your equation.
a graph of your equation looks like this:
you can see that the graph crosses the x-axis at x = -5 and at x = 2
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