Find the x-intercepts for the parabola: y = x^2 + 4x - 3

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The question asks for the x-intercepts of the parabola y = x^2 + 4x -3.

We will use the method of completing the square to find the x-intercepts.

=> 0= x^2 +4x -3.

=> 3 = x^2 + 4x

=> 3 + 4 = x^2 + 4x + 4

=> 7 = (x + 2)^2

=> -2 + sqrt 7 = x  or -2 - sqrt 7 = x

=>  The x intercepts are -2 + sqrt 7 and -2 - sqrt 7.

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The x-intercepts for a function are the points where the graph of the function meets the x-axis. Here the value of f(x) is 0.

For the given equation of the parabola: y = x^2 + 4x - 3, we find the x-intercepts by equating y = 0 and solving the resulting equation.

y = x^2 + 4x - 3 = 0

x1 = [-b + sqrt (b^2 - 4ac)]/2a

=> x1 = [ -4 + sqrt ( 16 + 12)]/2

=> -2 + sqrt 28 / 2

=> -2 + sqrt 7

x2 = -2 - sqrt 7

The x-intercepts of the parabola are at the points ( -2 + sqrt 7 , 0) and ( -2 - sqrt 7 , 0)

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