Question

acha as rais das equacoes 
a)x²-8x+7=0
b)3x²-7x+4=0
c)9y²-12y+4=0
d)9x²-24x+16=0
e)3x²-15x+12=0
f)x²-10x+25=0

Answer 1

a)
$x^{2} - 8x + 7 = 0$

$x = \frac{-b \pm \sqrt{-b^2 -4*a*c}}{2*a}$

a=1, b=−8, c=7
Δ=b2−4ac
Δ=(−8)2−4*(1)*(7)
Δ=64−28
Δ=36

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-(-8) \pm \sqrt{36}}{2*1} \\ \\ x = \frac{8 \pm 6}{2} \\ \\ x' = \frac{8 + 6}{2} \\ \\ x' = \frac{14}{2} \\ \\ x' = 7 \\ \\ \\ x'' = \frac{8 - 6}{2} \\ \\ x'' = \frac{2}{2} \\ \\ x'' = 1$

S= (7, 1)
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b)
$3x^{2} -7x+4=0$



a=3, b=−7, c=4
Δ=b2−4ac
Δ=(−7)2−4*(3)*(4)
Δ=49−48
Δ=1
$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-(-7) \pm \sqrt{1}}{2*3} \\ \\ x = \frac{7 \pm {1}}{6} \\ \\ x' = \frac{7 + {1}}{6} \\ \\ x' = \frac{8}{6} \\ \\ x' = \frac{4}{3} \\ \\ \\ x'' = \frac{7 - {1}}{6} \\ \\ x'' = \frac{6}{6} \\ \\ x'' = 1$

S ={$\frac{4}{3} , 1$}
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c) $9y^2-12y+4=0$

a=9, b=−12, c=4
Δ=b2−4ac
Δ=(−12)2−4*(9)*(4)
Δ=144−144
Δ=0

$y = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ y = \frac{-(-12) \pm \sqrt{0}}{2*9} \\ \\ y = \frac{12 \pm 0}{18} \\ \\ y' = \frac{12 + 0}{18} \\ \\ y' = \frac{12}{18} \\ \\ y' = \frac{2}{3} \\ \\ \\ y'' = \frac{12 - 0}{18} \\ \\ y'' = \frac{12}{18} \\ \\ y'' = \frac{2}{3}$

S = {$\frac{2}{3}$}
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d) $9 x^{2}-24x+16=0$

a=9, b=−24, c=16
Δ=b2−4ac
Δ=(−24)2−4*(9)*(16)
Δ=576−576
Δ=0

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-(-24) \pm \sqrt{0}}{2*9} \\ \\ x = \frac{24 \pm {0}}{18} \\ \\ x' = \frac{24 + {0}}{18} \\ \\ x' = \frac{24}{18} \\ \\ x' = \frac{4}{3} \\ \\ \\ x'' = \frac{24 - {0}}{18} \\ \\ x'' = \frac{24}{18} \\ \\ x'' = \frac{4}{3}$

S = {$\frac{4}{3}$}
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e) $3x^{2}-15x+12=0$

a=1, b=−5, c=4
Δ=b2−4ac
Δ=(−5)2−4*(1)*(4)
Δ=25−16
Δ=9

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-(-5) \pm \sqrt{9}}{2*1} \\ \\ x = \frac{5 \pm 3}{2} \\ \\ x' = \frac{5 + 3}{2} \\ \\ x' = \frac{8}{2} \\ \\ \ x' = 4 \\ \\ \\ x'' = \frac{5 - 3}{2} \\ \\ x'' = \frac{2}{2} \\ \\ \ x'' = 1$

S = {4, 1}
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f) $x^{2}-10x+25=0$

a=1, b=−10, c=25
Δ=b2−4ac
Δ=(−10)2−4*(1)*(25)
Δ=100−100
Δ=0
$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-(-10) \pm \sqrt{0}}{2*1} \\ \\ x = \frac{10 \pm \sqrt{0}}{2} \\ \\ x' = \frac{10 +0}{2} \\ \\ x' = \frac{10}{2} \\ \\ x' = 5 \\ \\ \\ x'' = \frac{10 -0}{2} \\ \\ x'' = \frac{10}{2} \\ \\ x'' = 5$

S= {5}