Question

sendo f(x)=x²-5x+6,determine:a)f(√3);
b)o valor de x para f(x)=21

Answer 1

$f(x)=x^2-5x+6$

 a - $f(\sqrt{3})=(\sqrt{3})^2-5(\sqrt{3})+6$
$f(\sqrt{3})=3-5\sqrt{3}+6$
$f(\sqrt{3})=9-5\sqrt{3}$
$f(\sqrt{3})\approx9-5*1,732$
$f(\sqrt{3})\approx9-8,66$
$f(\sqrt{3})\approx0,34$

b - $f(x)=x^2-5x+6$
$f(x)=21$
$x^2-5x+6=21$
$x^2-5x+6-21=0$
$x^2-5x-15=0$

$x=\frac{-b\pm\sqrt{\Delta}}{2a}$
$x=\frac{-(-5)\pm\sqrt{\Delta}}{2*1}$
$x=\frac{5\pm\sqrt{\Delta}}{2}$

$\Delta=b^{2}-4ac$
$\Delta=(-5)^{2}-4*1*(-15)$
$\Delta=25+60$
$\Delta=85$

$x=\frac{5\pm\sqrt{85}}{2}$