Question

determine  os  numeros  das  seguintes  equaçoes  a) (x+7).(x-7)+61=7x     letra  b)  3x   ao  quadrado menos  2x =2-x

Answer 1

 (x+7).(x-7)+61=7x  

 

x²-49+61=7x

 

x²-7x+12=0

 

(x-3)(x-4)=0

 

x=3 ou x=4

 

3x²-2x =2-x

3x²-x-2=0

 

delta=1-4.a.c

delta=1-4.3.-2

delta=1+24=25

raizdelta=5

 

x1=(-b+raizdelta)/2.a

x1=(1+5)/2.3

x1=6/6=1

 

x2=(-b-raizdelta)/2.a

x2=(1-5)/2.3

x2=-2/3

Answer 2

$<var>a)(x+7).(x-7)+61=7x\\x ^2-7x+7x+61=7x\\x^2_{a}-7x_{b}+12_{c}=0\\(Bhaskara)\\x=\frac{-b_+_-\sqrt{b^2-4ac}}{2a}\\\\x=\frac{-(-7)_+_-\sqrt{(-7)^2-4(1).(12)}}{2.1}\\\\x=\frac{7_+_-\sqrt{49-48}}{2}=>\frac{7_+_-1}{2}\\\\x^|=\frac{8}{2}=4\\x^{||}=\frac{6}{2}=3\\\\\\\\ b)3x^2-2x=2-x\\3x^2-2x-2+x=0\\3x^2_a-x_b-2_c=0\\(Bhaskara)\\\\x=\frac{-(-1)_+_-\sqrt{(-1)^2-4.(3).(-2)}}{2.3}=>\frac{1_+_-\sqrt{25}}{6}=>\frac{1_+_-5}{6}\\\\x^|=\frac{6}{6}=1\\x^{||}=-\frac{4}{6}=-\frac{2}{3}\\</var>$