Question

1) USANDO A FORMULA DE BHASKARA RESOLVA

A) x2+10x+16=0

 

2) resolva as equações 

a)x(x-1)=2

      3

 

b) x + x = x(x+1)

    4    2

 

c) x+x=6x²+1

    3  2     6

 

d) 0,1x² -1,5x + 5,6=0

 

 

 

Answer 1

a)$x^{2}+10x+16 =0 \\\\ \Delta = b^{2} - 4 \cdot a \cdot c \\\\ \Delta=(10)^{2} - 4 \cdot (1) \cdot (16) \\\\ \Delta =100-64 \\\\\ \Delta = 36 \\\\\\ x = \frac{-b \pm \sqrt{\Delta}}{2a} \\\\ x = \frac{-10 \pm \sqrt{36}}{2 \cdot 1} \\\\ x = \frac{-10 \pm 6}{2} \\\\\\ \Rightarrow x'=\frac{-10 + 6}{2} = \frac{-4}{2} = \boxed{-2} \\\\ \Rightarrow x''=\frac{-10 - 6}{2} = \frac{-16}{2}= \boxed{-8} \\\\\\ \boxed{\boxed{S=\{-2,-8 \}}}$



2-a) $\frac{x(x-1)}{3}= 2 \\\\ \frac{x^{2}-x}{3}=2 \\\\ \text{MMC=3} \\\\ \frac{x^{2}-x}{3}=\frac{2^{\times 3}}{1^{\times 3}} \\\\ \frac{x^{2}-x}{\not{3}}=\frac{6}{\not{3}} \\\\ x^{2}-x=6 \\\\ x^{2}-x-6= 0$

$\Delta = b^{2} - 4 \cdot a \cdot c \\\\ \Delta = (-1)^{2} - 4 \cdot (1) \cdot (-6) \\\\ \Delta =1+24 \\\\ \Delta =25 \\\\\\ x =\frac{-b \pm \sqrt{\Delta}}{2a} \\\\ x =\frac{-(-1) \pm \sqrt{25}}{2 \cdot 1} \\\\ x =\frac{1 \pm 5}{2} \\\\\\ \Rightarrow x' =\frac{1 + 5}{2} = \frac{6}{2} = \boxed{3} \\\\ \Rightarrow x'' = \frac{1 - 5}{2}= \frac{-4}{2}= \boxed{-2} \\\\\\ \boxed{\boxed{S =\{-2,3\}}}$



b) $\frac{x}{4} + \frac{x}{2} = x(x+1) \\\\ \frac{x}{4} + \frac{x}{2} = x^{2}+x \\\\ \text{MMC = 4} \\\\\ \frac{x}{4} + \frac{x^{\times 2}}{2^{\times2}} = \frac{(x^{2}+x)^{\times 4}}{1^{\times 4}} \\\\ \frac{x}{\not{4}} + \frac{2x}{\not{4}} = \frac{4x^{4}+4x}{\not{4}} \\\\ x+2x = 4x^{2}+4x \\\\ 4x^{2}+4x-2x-x = 0 \\\\ 4x^{2}+x = 0$

$\Delta = b^{2} - 4 \cdot a \cdot c \\\\ \Delta = (1)^{2} - 4 \cdot (4) \cdot (0) \\\\ \Delta = 1-0 \\\\ \Delta = 1 \\\\\\ x =\frac{-b \pm \sqrt{\Delta}}{2a} \\\\ x =\frac{-1 \pm \sqrt{1}}{2 \cdot 4} \\\\ x =\frac{-1 \pm 1}{8} \\\\\\ \Rightarrow x' = \frac{-1 + 1}{8} = \frac{0}{8} = \boxed{0} \\\\\\ \Rightarrow x'' =\frac{-1 - 1}{8} = \frac{-2}{8} = \boxed{\frac{-1}{4}} \\\\\\ \boxed{\boxed{S = \{0, -\frac{1}{4}\}}}$



c) $\frac{x}{3} + \frac{x}{2} = \frac{6x^{2}+1}{6} \\\\ \text{MMC = 6} \\\\ \frac{x^{\times 2}}{3^{\times 2}} + \frac{x^{\times 3}}{2^{\times 3}} = \frac{6x^{2}+1}{6} \\\\ \frac{2x}{\not{6}} + \frac{3x}{\not{6}} = \frac{6x^{2}+1}{\not{6}} \\\\ 2x+3x = 6x^{2} + 1 \\\\ 6x^{2}-2x-3x+1 = 0 \\\\ 6x^{2} - 5x+1 = 0$

$\Delta = b^{2} - 4 \cdot a \cdot c \\\\ \Delta = (-5)^{2} - 4 \cdot (6) \cdot (1) \\\\ \Delta = 25-24 \\\\ \Delta = 1 \\\\\\ x = \frac{-b \pm \sqrt{\Delta}}{2a} \\\\ x = \frac{-(-5) \pm \sqrt{1}}{2 \cdot 6} \\\\ x = \frac{5 \pm 1}{12} \\\\\\ \Rightarrow x' = \frac{5 + 1}{12} = \frac{6}{12} = \boxed{\frac{1}{2}} \\\\ \Rightarrow x'' = \frac{5 - 1}{12} = \frac{4}{12} = \boxed{\frac{1}{3}} \\\\\\ \boxed{\boxed{S = \{\frac{1}{3}, \frac{1}{2}\}}}$



d) Antes da fazermos, vamos transformar tudo em fração para facilitar os cálculos:

$0,1 = \frac{1}{10} \\\\ 1,5 = \frac{15}{10} \\\\ 5,6 = \frac{56}{10}$


$\frac{1x^{2}}{10}-\frac{15x}{10} + \frac{56}{10} = 0 \\\\ \frac{1x^{2}}{\not{10}}-\frac{15x}{\not{10}} + \frac{56}{\not{10}} = 0 \\\\ x^{2}-15x+56 = 0 \\\\ \Delta = b^{2} - 4 \cdot a \cdot c \\\\ \Delta = (-15)^{2}- 4 \cdot (1) \cdot (56) \\\\ \Delta = 225-224 \\\\ \Delta = 1 \\\\\\ x = \frac{-b \pm \sqrt{\Delta}}{2a} \\\\ x = \frac{-(-15) \pm \sqrt{1}}{2 \cdot 1} \\\\\ x = \frac{15 \pm 1}{2}$


$\Rightarrow x' = \frac{15 + 1}{2} = \frac{16}{2} = \boxed{8} \\\\ \Rightarrow x'' = \frac{15 - 1}{2} = \frac{14}{2} = \boxed{7} \\\\\\ \boxed{\boxed{S = \{7,8\}}}$

Answer 2

x = -b ± √ b² - 4ac
      --- ----------
           2a

a) x²+10x+16=0

x =    -10  ± √100 - 4 . 16        
                  2
x =    -10  ± √36             
                  2

x' = -10 + 6 /2 = -2
x" = -10 - 6/2 = -8

2)
a)x(x-1)=2     
        3
x (x - 1) = 2 .3
x² - x = 6
x² - x - 6 = 0

x =    1  ± √1 - 4 . (-6)           
              2
x =    1  ± √25             
                2
x'= 1 + 5/2  =  3
x"= 1 - 5/2  =   -2
 
b) x + x = x(x+1)   
    4    2 
 x+ 2x = x² + x
     4
x + 2x = 4 (x² + x)
3x = 4x² + 4x
4x² + 7x = 0

x =    -7  ±√ 49 - 4.4.0          
             8
x =    -7  ±√ 49          
             8
x' = 0 
x" = -14/8 = -7/4

c) x+x=6x²+1        (tira o mmc e corta o denominador porque é uma equação
    3  2     6                                      permite fazer isso)

2x + 3x = 6x² + 1

6x² - 5x + 1 = 0
x =   5 ± √ 25  - 4. 6            
               12 
x =   5 ± √ 1           
               12 
x' = 1/2
x" = 1/3 


d) 0,1x² -1,5x + 5,6=0
x = 1,5 ± √2,25 - 4. 0,1 . 5,6
           0,2
x = 1,5 ± √0,01
           0,2
x' = 1,5 + 0,1/2
x' = 8

x" = 1,5 - 0,1 /2
x" = 0,7

Espero ter te ajudado, desculpe pela demora (: