Question

What is the solution of StartRoot x + 12 EndRoot = x?

Answer 1

Answer:

B)  x=4

Step-by-step explanation:

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Answer 2

Answer:

x = 4      or x = -3

Step-by-step explanation:

$\sqrt{x+12}=x$

Take square both the sides,

$(\sqrt{x+12})^{2}=x^{2}\\\\x+12=x^{2}$

x² - x - 12 = 0

x² - 4x + 3x - 3*4 =0

x*(x - 4) + 3*(x - 4)= 0

(x- 4 )(x+3) =0

x -4 =0   or x + 3 = 0

x = 4      or x = -3