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How to solve this quadratic equation x^2-4x-5=0

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2 Answers

 The goal here is to solve for the unknown variable, x, when the equation is equal to zero. Since the equation is a 2nd degree polynomial function (or a quadratic function), we first want to change it into a product of two binomials by factoring. Once we have done that, we can apply the zero product property to solve for x.    

        x2 - 4x - 5 = 0

    (x - 5)(x + 1) = 0

          x - 5 = 0                 and               x + 1 = 0

            x = 5                                             x = -1

There are two ways you could do this: factoring and then using the zero product rule, or using the quadratic formula.

The Quadratic Formula
x = [-b ± √(b2 - 4ac)]/2a, where your original quadratic equation takes the form ax2 + bx + c = 0
In your problem, a = 1, b = -4, and c = -5
When you plug these values into the formula and solve for x, you will get two different answers -- one for adding in the numerator, and one for subtracting in the numerator.

 

Zero Product Rule
When you have (x+a)(x+b)=0, your two solutions will be a and b.
One of these must be true in order for the product to equal 0 (Zero Product Rule): x+a=0, OR x+b=0
To factor your equation to make it look like that, find two things that multiply to equal -5, and add up to -4: -5 and 1.
Now factor your equation: x2 - 4x - 5 = 0 ===> (x-5)(x+1) = 0
x-5=0  ===> x = 5
x+1=0 ===> x =  -1