Question

subtraindo- se 4 de um certo número, obtém - se triplo da sua raiz quadrada. Então este número é :


A) 1

B) 4

C) 9

D) 16

Answer 1

Seja x o número procurado, logo

$x-4=3\cdot \sqrt{x}\\\\(x-4)^2=(3\cdot \sqrt{x})^2\\\\(x-4)^2=9x\\\\x^2-8x+16-9x=0\\\\x^2-17x+16=0\\\\a=1\ \ \ \ b=-17\ \ \ \ c=16\\\\\triangle=b^2-4ac\\\triangle=(-17)^2-4\cdot(1)\cdot(16)\\\triangle= 289- 64\\\triangle=225\\\\\\x= \dfrac{-b \frac{+}{-} \sqrt{\triangle} }{2a}\\\\\\x= \dfrac{-(-17) \frac{+}{-} \sqrt{225} }{2} \\\\\\x= \dfrac{17 \frac{+}{-}15 }{2}\\\\\\x'= \dfrac{32}{2}\\\\\\x'=16\\\\\\x"= \dfrac{2}{2}\\\\\\x"=1$

Checando

$Para\ x=16\\\\16 - 4 = 3\cdot \sqrt{16}\\\\12=3\cdot4\\\\12=12\ \ \ \checkmark\\\\\\Para\ x=1\\\\1-4=3\cdot \sqrt{1}\\\\-3=3\ \ \ \ \ \ X$

Letra D