Question

escreve as equações na forma geral e resolva em IR

a)x²+3=4x

b)x(x+2)=3

c)x(2x-1)+6=4(x+1)$<var>x²+3=4x </var>$

Answer 1

 x² - 4x + 3 = 0
► FORMULA DE BASKHARA: 
►
► ax² + bx + c = 0
►
► a = 1
► b = -4
► c = 3
►
► ∆² = b² - 4ac = (-4)² - 4.1.3 = 16 -12 = 4
► ∆ = 2
►
► x' = -b/2a + ∆/2a = 4/2 + 2/2= 3
► x" = -b/2a - ∆/2a = 4/2 - 2/2 = 1

Answer 2

a) $\text{x}^2+3=4\text{x}$

 

Temos que:

 

$\text{x}^2+3=4\text{x}$

 

$\text{x}^2-4\text{x}+3=0$

 

$\text{x}=\dfrac{-(-4)\pm\sqrt{(-4)^2-4\cdot1\cdot3}}{2\cdot1}=\dfrac{4\pm2}{2}$

 

Logo, as raízes são:

 

$\text{x}'=\dfrac{4+2}{2}=3$

 

$\text{x}''=\dfrac{4-2}{2}=1$

 

 

 

b) $\text{x}\cdot(\text{x}+2)=3$

 

Temos que:

 

$\text{x}^2+2\text{x}-3=0$

 

$\text{x}=\dfrac{-2\pm\sqrt{2}^2-4\cdot1\cdot(-3)}}{2\cdot1}+\dfrac{-2\pm4}{2}$

 

Desta maneira, as raízes são:

 

$\text{x}'=\dfrac{-2+4}{2}=1$

 

$\text{x}"=\dfrac{-2-4}{2}=-3$

 

 

 

c) $\text{x}(2\text{x}-1)+6=4\cdot(\text{x}+1)$

 

$2\text{x}^2-\text{x}+6=4\text{x}+4$

 

$2\text{x}^2-5\text{x}+2=0$

 

$\text{x}=\dfrac{-(-5)\pm\sqrt{(-5)^2-4\cdot2\cdot2}}{2\cdot2}=\dfrac{5\pm3}{4}$

 

Logo, as raízes são:

 

$\text{x}'=\dfrac{5+3}{4}=2$

 

$\text{x}"=\dfrac{5-3}{4}=\dfrac{2}{4}=\dfrac{1}{2}$