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If 2x^2 + 6x = 36, what are the possible values of x ?

Sagot :

jaccusdarrisaw55

Answer:

f(x) = 2(x + 3)(x - 6)

Step-by-step explanation:

xero099

The possible values of [tex]x[/tex] for [tex]2\cdot x^{2}+6\cdot x = 36[/tex] are [tex]x_{1} = 3[/tex] and [tex]x_{2} = -6[/tex].

The expression [tex]2\cdot x^{2}+6\cdot x = 36[/tex] is equivalent to the expression [tex]2\cdot x^{2}+6\cdot x -36 = 0[/tex], which represents a second order polynomial, that is, a polynomial of the form:

[tex]y = a\cdot x^{2}+b\cdot x + c[/tex] (1)

Where [tex]a,\,b,\,c[/tex] are coefficients.

The possible values of x represents all possible roots of x, which are found analytically by the quadratic formula:

[tex]x = \frac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}[/tex] (2)

If we know that [tex]a = 2[/tex], [tex]b = 6[/tex] and [tex]c = -36[/tex], then all possible values of [tex]x[/tex] are, respectively:

[tex]x_{1} = 3[/tex], [tex]x_{2} = -6[/tex]

The possible values of [tex]x[/tex] for [tex]2\cdot x^{2}+6\cdot x = 36[/tex] are [tex]x_{1} = 3[/tex] and [tex]x_{2} = -6[/tex].

We kindly invite to check this question on second order polynomials: https://brainly.com/question/24356198

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