Question

What is the solution of the equation?

Answer 1

Answer: 7

Step-by-step explanation:

$\sqrt{2x-5}+4 = x$

$\sqrt{2x-5} = x-4$   subtracted 4 from both sides

$(\sqrt{2x-5})^2 = (x-4)^2$   squared both sides to eliminate square root

2x - 5 = x² - 8x + 16         expanded right side

       0 = x² - 10x + 21        subtracted 2x and added 5 on both sides

        0 = (x - 3) (x - 7)        factored right side

0 = x - 3     0 = x - 7          applied zero product property

  x = 3          x = 7             solved for x

Check:

x = 3

$\sqrt{2(3)-5}+4 = (3)$

$\sqrt{1}+4 = 3$

1 + 4 = 3  

FALSE!  x = 3 is NOT a valid solution

x = 7

$\sqrt{2(7)-5}+4 = (7)$

$\sqrt{9}+4 = 7$

3 + 4 = 7  

TRUE! x = 7 IS a valid solution

Answer 2

$\sqrt{2x - 5} = x - 4$
$2x - 5 = (x - 4) {}^{2}$
$2x - 5 = x {}^{2} - 8x + 16$
$x {}^{2} - 10x + 21 = 0$
$(x - 3)(x - 7) = 0$
$x = 3 \\ x = 7$