Question

Escreva as equações abaixo na forma reduzida e resolva-as:

A) x(x+3) = 5x+15

B) 3y+1/2 = y²-1/3

C) (x+4)² = 9x+22

D) (x-1)²+3x = x+26

E) (x+4).(x-1) = 5x+20

Answer 1

$A) x(x+3) = 5x+15 x^{2}+3x =5x+15 x^{2} +3x-5x-15 = 0 x^{2}-2x-15 = 0$

$\frac{-(-2)+- \sqrt{(-2)^{2}-4.1.-15 } }{2.1}$
$\frac{2+- \sqrt{4+60} }{2}$
$\frac{2+- \sqrt{64} }{2}$
$\frac{2+-8}{2}$

$x^{'} = \frac{2+8}{2} = \frac{10}{2} = 5 x^{''}= \frac{2-8}{2} = \frac{-6}{2} = -3$

$S= [5,-3]$


$B) \frac{3y+1}{2} = \frac{ y^{2}-1 }{3}$

$(3y+1).3 = ( y^{2}-1).2 9y+3 = 2y^{2}-2 9y+3-2y^{2}+2 = 0 -2y^{2} +9y+5 = 0$

$\frac{-9+- \sqrt{ 9^{2}-4.-2.5 } }{2.-2}$
$\frac{-9+- \sqrt{81+40} }{-4}$
$\frac{-9+- \sqrt{121} }{-4}$
$\frac{-9+-11}{-4}$

$x^{'} = \frac{-9+11}{-4} = \frac{2}{-4} = \frac{1}{-2} x^{''} = \frac{-9-11}{4} = \frac{-20}{-4} = 5$

$S= [- \frac{1}{2},5 ]$


$C) (x+4)^{2} =9x+22$

$(x+4).(x+4) = 9x+22 x^{2} +4x+4x+16 = 9x+22 x^{2} +8x +16 -9x-22 = 0 x^{2} -x-6 = 0$

$\frac{-(-1)+- \sqrt{(-1)^{2}-4.1.-6} }{2.1}$
$\frac{1+- \sqrt{1+24} }{2}$
$\frac{1+- \sqrt{25} }{2}$
$\frac{1+-5}{2}$

$x^{'}= \frac{1+5}{2} = \frac{6}{2} = 3 x^{''}= \frac{1-5}{2} = \frac{-4}{2} = -2$

$S= [3,-2]$


$D) (x-1)^{2}+3x =x+26$

$(x-1).(x-1)+3x = x+26 x^{2} -x-x+1+2x = x+26 x^{2} -x-x+1+3x-x-26 = 0 x^{2} -25=0$

$x^{2} =25 x= +- \sqrt{25} x= +-5$

$S= [-5,5]$


$E) (x+4).(x-1) = 5x+20$

$x^{2} -x+4x-4 = 5x+20 x^{2} -5x-x+4x-4-20 = 0 x^{2} -2-24 = 0$

$\frac{-(-2)+- \sqrt{(-2)^{2}-4.1.-24} }{2.1}$
$\frac{2+- \sqrt{4+96} }{2}$
$\frac{2+- \sqrt{100} }{2}$
$\frac{2+-10}{2}$

$x^{'} = \frac{2+10}{2} = \frac{12}{2} = 6 x^{''} = \frac{2-10}{2} = \frac{-8}{2} = -4$

$S = [6, -4]$

Answer 2

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