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Sagot :
Answer:
For [tex]x = -2[/tex]
Step-by-step explanation:
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]
[tex]\Delta = b^{2} - 4ac[/tex]
In this question:
[tex]f(x) = x^2 - 2x[/tex]
[tex]g(x) = 6x + 4[/tex]
[tex]f(x) + g(x) = (f+g)(x) = x^2 - 2x + 6x + 4 = x^2 + 4x + 4[/tex]
Which is a value of x does (f+g)(x) =0
[tex]x^2 + 4x + 4[/tex]
Quadratic equation with [tex]a = 1, b = 4, c = 4[/tex]
So
[tex]\Delta = 4^{2} - 4*1*4 = 0[/tex]
[tex]x_{1} = \frac{-4 + \sqrt{0}}{2} = -2[/tex]
[tex]x_{2} = \frac{-4 - \sqrt{0}}{2} = -2[/tex]
For [tex]x = -2[/tex]
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