Question

$\sqrt{x^{2}-16 }-3\sqrt{x-4}=0$

Answer 1

Answer:

$\tt{} x_{1} = 4 \\ \tt{} x_{2} = 5$

Step-by-step explanation:

$\tt{} \sqrt{ {x}^{2} - 16 } - 3 \sqrt{x - 4} = 0$

$\tt{} \sqrt{ {x}^{2} - 16 } = 3 \sqrt{x - 4}$

$\tt{}{x}^{2} - 16 = 9(x - 4)$

$\tt{}(x - 4).(x + 4) = 9(x - 4)$

$\tt{}(x - 4) \times (x + 4 - 9) = 0$

$\tt{}( x - 4) \times (x - 5) = 0$

$\tt{}x - 4 = 0 \\ \tt{}x - 5 = 0$

$\tt{}x = 4 \\ \tt{}x = 5$

$\tt{} x_{1} = 4 \\ \tt{} x_{2} = 5$

Answer 2

Answer:

x = 4, 5.

Step-by-step explanation:

√(x^2 - 16) - 3√(x - 4) = 0

√(x^2 - 16) = 3√(x - 4)

Square both sides of the equation:

x^2 - 16 = 9(x - 4)

x^2 - 9x - 16 + 36 = 0

x^2 - 9x + 20 = 0

(x - 4)(x - 5) = 0

x = 4, 5.

As the original equation contained square roots:

We test for extraneous solutions:

x = 4:

√(4^2 - 16) - 3√(4 - 4) = 0 - 3*0 = 0  so x = 4 is OK.

x = 5:

√(5^2 - 16) - 3√(5 - 4) = √9 - 3√1 = 0  so x = 5 is also OK.