Question

determine as raízes das equações a seguir
a)x²-5x+6=0
b)2y²+3y-14=0
c)(2x+3) (x-1)=3​

Answer 1

Olá!

Questão a)

$x^{2} -5x+6=0~~~\to ~~~a=1~,~b=-5~,~c=6\\ \\ \\ \\ \dfrac{-b\pm\sqrt{b^{2} -4\cdot a\cdot c} }{2\cdot a}~~\to ~~\dfrac{-(-5)\pm\sqrt{(-5)^{2} -4\cdot 1\cdot 6} }{2\cdot 1}~~\to ~~\dfrac{5\pm\sqrt{25-24} }{2}~~\to ~~\dfrac{5\pm1}{2} \\ \\ \\ \\ \boxed{x'=3~~~,~~~x''=2}$

--------------------------------------------------------------------------------------------------------------

Questão b)

$2y^{2} +3y-14=0~~~\to ~~~a=2~,~b=3~,~c=-14\\ \\ \\ \\ \dfrac{-b\pm\sqrt{b^{2} -4\cdot a\cdot c} }{2\cdot a}~~\to ~~\dfrac{-3\pm\sqrt{3^{2} -4\cdot 2\cdot (-14)} }{2\cdot 2}~~\to ~~\dfrac{-3\pm\sqrt{9+112} }{4}~~\to ~~\dfrac{-3\pm\sqrt{121} }{4} \\ \\ \\ \\ \dfrac{-3\pm11}{4} ~~~\to ~~~ \boxed{y'=2~~~,~~~y''=-\dfrac{7}{2} }$

--------------------------------------------------------------------------------------------------------------

Questão c)

$(2x+3)\cdot(x-1)=3\\ \\ 2x^{2} -2x+3x-3=3\\ \\ 2x^{2} -2x+3x-3-3=0\\ \\ \\ \\ 2x^{2} +x-6=0~~~\to ~~~a=2~,~b=1~,~c=-6$

$\dfrac{-b\pm\sqrt{b^{2} -4\cdot a\cdot c} }{2\cdot a}~~\to ~~\dfrac{-1\pm\sqrt{1^{2} -4\cdot 2\cdot (-6)} }{2\cdot 2}~~\to ~~\dfrac{-1\pm\sqrt{1+24} }{4}~~\to ~~\dfrac{-1\pm\sqrt{25} }{4} \\ \\ \\ \\ \dfrac{-1\pm5}{4} ~~~\to ~~~ \boxed{x'=1~~~,~~~x''=-\dfrac{3}{2} }$

:)