Question

tópicos de cálculo
Função do segundo grau
Determine as raízes (zeros) reais de cada uma das funções seguintes
a) y=2x²-3x+1
b) y=4x-x²
c)y=-x²+2x+15
d)y=9x²-1
e)y=-x²+6x-9
f)y=x³-3∛3.x+6

Answer 1

encontrar as raízes de uma equação do segundo grau..
vc pode usar bhaskara..pode fatorar..usar soma e produto
************************************************************************************************
a)$y=2x^2-3x+1\\\\0=2x^2-3x+1$

a = 2 (porque acompanha o x²)
b = -3 (porque acompanha o x)
c = 1 (é o termo independente )
utilizando bhaskara
$\frac{-b\pm \sqrt{b^2-4*a*c} }{2*a} = \frac{-(-3)\pm \sqrt{(-3)^2-4*2*1} }{2*2} = \frac{3\pm \sqrt{9-8} }{4} \\\\ \frac{3\pm \sqrt{1} }{4} = \frac{3\pm1}{4} \\\\x'= \frac{3+1}{4} =1\\\\x''= \frac{3-1}{4}= \frac{2}{4} = \frac{1}{2}$

isso significa que quando x for = 1 ou 1/2 
y vai ser =0
********************************************************************************************************
b) $y=4x-x^2\\\\0=4x-x^2\\\\0+x^2=4x\\\\ \frac{x^2}{x}=4\\\\x=4$

uma das raizes é 4 
e a outra raíz é 0 ..porque os dois termos dependem de x..então se vc substituir
x por 0
y será igual a 0

se vc fizer utilizando bhaskara o resultado sera o mesmo
*******************************************************************************************************
$c)y=-x^2+2x+15$

a = -1 (porque acompanha o x²)
b = 2 (porque acompanha o x)
c = 15 (termo que não depende de x)

$\frac{-b\pm \sqrt{b^2-4*a*c} }{2*a} =\frac{-2\pm \sqrt{2^2-4*(-1)*15} }{2*(-1)} = \frac{-2\pm \sqrt{4+60} }{-2} \\\\= \frac{-2\pm \sqrt{64} }{-2} = \frac{-2\pm8}{-2} \\\\x'= \frac{-2+8}{-2} = \frac{6}{-2} =-3\\\\x''= \frac{-2-8}{-2}= \frac{-10}{-2} =5$
*********************************************************************************************************
d)$y=9x^2-1\\\\0=9x^2-1\\\\0+1=9x^2\\\\ \frac{1}{9} =x^2\\\\ \pm \sqrt{ \frac{1}{9} } \\\\x'=\sqrt{ \frac{1}{9} } \\\\x''=-\sqrt{ \frac{1}{9} }$
********************************************************************************************************
$y=-x^2+6x-9$
$\frac{-b\pm \sqrt{b^2-4*a*c} }{2*a}=\frac{-6\pm \sqrt{6^2-4*(-1)*-9} }{2*-1}= \frac{-6\pm0}{-2} \\\\x'=x''= \frac{-6}{2} =-3$

*******************************************************************************************
$f)y=x^2-3 \sqrt{3}x +6$

a = 1
b = -3√3 (porque acompanha o x)
c = 6

$\frac{-b\pm \sqrt{b^2-4*a*c} }{2*a}=\frac{-(-3 \sqrt{3}) \pm \sqrt{(-3 \sqrt{3} )^2-4*1*6} }{2*1}\\\\ \frac{3 \sqrt{3} \pm \sqrt{((-3)^2*( \sqrt{3})^2-24 } }{2} = \frac{3 \sqrt{3} \pm \sqrt{9*3-24 } }{2} \\\\ = \frac{3 \sqrt{3} \pm \sqrt{27-24 } }{2} = \frac{3 \sqrt{3}\pm \sqrt{3} }{2} \\\\\boxed{x'= \frac{3 \sqrt{3}+ \sqrt{3} }{2} = \frac{4 \sqrt{3} }{2} = 2 \sqrt{3} }} \\\\\\\boxed{x''= \frac{3 \sqrt{3}- \sqrt{3} }{2}= \frac{2 \sqrt{3} }{2} = \sqrt{3} }$