Question

Utilizando os conhecimentos de equação do 2°grau,determine o valor de x,sabendo que a fórmula da área de um triângulo é:A=(b.h)/2

Answer 1

a)

x(x - 7)/2 = 60, para x = 15

15(15 - 7)/2 = 60

15 . 8/2 = 60

15 . 4 = 60

60 = 60

b)

(x - 18)(x - 1)/2 = 84, para x = 25

(25 - 18)(25 - 1)/2 = 84

7 . 24/2 = 84

7 . 12 = 84

84 = 84

Resposta:

a) x = 15

b) x = 25

Answer 2

Explicação passo-a-passo:

a)

$\sf A=\dfrac{b\cdot h}{2}$

$\sf \dfrac{x\cdot(x-7)}{2}=60$

$\sf x\cdot(x-7)=2\cdot60$

$\sf x^2-7x=120$

$\sf x^2-7x-120=0$

$\sf \Delta=(-7)^2-4\cdot1\cdot(-120)$

$\sf \Delta=49+480$

$\sf \Delta=529$

$\sf x=\dfrac{-(-7)\pm\sqrt{529}}{2\cdot1}=\dfrac{7\pm23}{2}$

$\sf x'=\dfrac{7+23}{2}~\Rightarrow~x'=\dfrac{30}{2}~\Rightarrow~\red{x'=15}$

$\sf x"=\dfrac{7-23}{2}~\Rightarrow~x"=\dfrac{-16}{2}~\Rightarrow~\red{x"=-8}$ (não serve)

Logo, x = 15

b)

$\sf A=\dfrac{b\cdot h}{2}$

$\sf \dfrac{(x-18)\cdot(x-1)}{2}=84$

$\sf (x-18)\cdot(x-1)=2\cdot84$

$\sf x^2-x-18x+18=168$

$\sf x^2-19x+18=168$

$\sf x^2-19x+18-168=0$

$\sf x^2-19x-150=0$

$\sf \Delta=(-19)^2-4\cdot1\cdot(-150)$

$\sf \Delta=361+600$

$\sf \Delta=961$

$\sf x=\dfrac{-(-19)\pm\sqrt{961}}{2\cdot1}=\dfrac{19\pm31}{2}$

$\sf x'=\dfrac{19+31}{2}~\Rightarrow~x'=\dfrac{50}{2}~\Rightarrow~\red{x'=25}$

$\sf x"=\dfrac{19-31}{2}~\Rightarrow~x"=\dfrac{-12}{2}~\Rightarrow~\red{x"=-6}$ (não serve)

Logo, x = 25