Question

Qual a conta completa de:

A) (x - 2)² = 25

B) (x + 3)² = 49

C) (8 - x)² = 1

D) (5x - 4)² = 9

E) (21x - 7)² = 0 ?




Eu realmente preciso de ajuda nessas contas, tenho de entregar o trabalho amanhã!

Answer 1

(x - 2)² = 25

(x - 2) . (x - 2) = 25

x² - 2x - 2x + 4 - 25 = 0

x² + 4x - 21 = 0

$\frac{-4+-\sqrt{4^{2}-4(1)(-21)}}{2(1)}$

$\frac{-4+-\sqrt{100}}{2}$

$x'=\frac{-4-10}{2} = -7$

$x''=\frac{-4+10}{2} = 3$

(x + 3)² = 49

(x + 3) . (x + 3) = 49

x² + 3x + 3x + 9 - 49 = 0

x² + 6x - 40 = 0

$\frac{-6+-\sqrt{6^{2}-4(1)(-40)}}{2(1)}$

$\frac{-6+-\sqrt{196}}{2}$

$x'=\frac{-6-14}{2} = -10$

$x'''=\frac{-6+14}{2} = 4$

(8 - x)² = 1

(8 - x) . (8 - x) = 1

64 - 8x - 8x + x² - 1 = 0

63 + 16x + x² = 0

$\frac{-16+-\sqrt{16^{2}-4(1)(63)}}{2(1)}$

$\frac{-16+-\sqrt{0}}{2}$

$x=\frac{-16}{2}=-8$

(5x - 4)² = 9

(5x - 4) . (5x - 4) = 9

25x² - 20x - 20x + 16 - 9 = 0

25x² + 40x + 7 = 0

$\frac{-40+-\sqrt{40^{2}-4(25)(7)}}{2(25)}$

$\frac{-40+-\sqrt{900}}{50}$

$x'=\frac{-40-30}{50} = -\frac{7}{5}$

$x''=\frac{-40+30}{50} = -\frac{1}{5}$

(21x - 7)² = 0

(21x - 7) . (21x - 7) = 0

441x² - 147x - 147x + 49 = 0

441x² + 294x + 49 = 0

$\frac{-294+-\sqrt{294^{2}-4(441)(49)}}{2(441)}$

$\frac{-294+-\sqrt{0}}{882}$$x=\frac{-294}{882} = -\frac{1}{3}$