Question

Resolva as equações  do 2º grau , usando a formula de bhaskara .

a: x² -12x + 36 = 0 
b: x² - x - 42 = 0
c: x² - 15x + 56 = 0
d: t² + 3t - 2 = 0
e: 49x² - 42 - 40 =0
f: 2x² - 128 = 0

Answer 1

$a)x^2 -12x+36=0$

Δ = $(-12)^2-4.1.36$
Δ = $144-144$
Δ = $0$

$X = \frac{-(-12)+- \sqrt{0} }{2.1} => X = \frac{12}{2} =6$

$b)x^2-x-42=0$

Δ = $(-1)^2-4.1.(-42)$
Δ = $1+168$
Δ = $169$

$X = \frac{-(-1)+- \sqrt{169} }{2.1} =>X= \frac{1+-169}{2}$

x' = $\frac{1+169}{2}=> \frac{170}{2} =85$                      x" = $\frac{1-169}{2} => \frac{-168}{2} =84$

$c) x^2-15x+56=0$

Δ = $(-15)^2-4.1.56$
Δ = $225-224$
Δ = $1$

$X = \frac{-(-15)+- \sqrt{1} }{2.1} => X=\frac{15+-1}{2}$

x' = $\frac{15+1}{2}=> \frac{16}{2} = 8$                x" = $\frac{15-1}{2} => \frac{14}{2}=7$

$d)t^2+3t-2=0$

Δ = $3^2-4.1.(-2)$
Δ = $9+8$
Δ = $17$

$T = \frac{-3+- \sqrt{17} }{2.1} => \frac{-3+-4,1}{2}$

t' = $\frac{-3+4,1}{2} => \frac{1,1}{2} = 0,55$ aproximadamente             t" = $\frac{-3-4,1}{2} => \frac{7,1}{2}= 3,55$

$e)49x^2-42x-40=0$

Δ = $(-42)^2-4.49.(-40)$
Δ = $1764+7840$
Δ = $9604$

$X = \frac{-(-42)+- \sqrt{9604} }{2.49} => X = \frac{42+-98}{98}$

x' = $\frac{42+98}{98} => \frac{140}{98} = 1,43$ aproximadamente           x" = $\frac{42-98}{98} => \frac{-56}{98} = 0,57$ aproximadamente.

$f)2x^2-128=0$

Não é necessário usar bháskara

$2x^2=128 \\ \\ x^2 = \frac{128}{2} \\ \\ x^2 = 64 \\ x = \sqrt{64} \\ x=8$