Question

se considera funcția f:R->R, f(x)=x^2-3.determinati numărul real a, știind ca f(a) =-2a ​

Answer 1

Răspuns:

a = -3; a = 1

Explicație pas cu pas:

$f(x) = {x}^{2} - 3$

$f(a) = {a}^{2} - 3$

${a}^{2} - 3 = - 2a \\ {a}^{2} + 2a - 3 = 0 \\ (a + 3)(a - 1) = 0$

$a + 3 = 0 \implies a = - 3$

$a - 1 = 0 \implies a = 1$

Answer 2

$f:\mathbb{R}\rightarrow \mathbb{R};~~f(x)=x^2-3;~~f(a)=-2a\Rightarrow a=?\\\\f(a)=-2a\Rightarrow a^2-3=-2a\Rightarrow a^2+2a-3=0\\\\\Delta=2^2-4\cdot1\cdot(-3)=4+12=16 > 0\\\\a_1=\dfrac{-2-\sqrt{16} }{2\cdot1}=\dfrac{-2-4}{2}=-\dfrac{6}{2} =-3 ~~~~~~~~~~~~~~~~~~~~~\boxed{a_1=-3}\\\\a_2=\dfrac{-2+\sqrt{16} }{2\cdot 1} =\dfrac{-2+4}{2}=\dfrac{2}{2}=1~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{a_2=1}$