Question

Solve the equation square root of x plus three plus four equals five for the variable. Show each step of your solution process

Answer 1

For this case we must solve the following equation:

$\sqrt {x+3}+4 = 5x$

We subtract 4 on both sides of the equation:

$\sqrt {x+3}+4-4 = 5x-4\\\sqrt {x+3} = 5x-4$

We square both sides of the equation:

$(\sqrt {x+3}) ^ 2 = (5x-4) ^ 2\\x+3 = (5x) ^ 2-2 (5x) (4)+4 ^ 2\\x+3 = 25x ^ 2-40x+16\\0 = 25x ^ 2-41x+13$

We have an equation of the form:

$ax ^ 2+bx+c = 0$

Where:

$a = 25\\b = -41\\c = 13$

The solutions are given by:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

Substituting:

$x = \frac {- (- 41) \pm \sqrt {(- 41) ^ 2-4 (25) (13)}} {2 (25)}\\x = \frac {41 \pm \sqrt {1681-1300}} {50}\\x = \frac {41 \pm \sqrt {381}} {50}$

So, the roots are:

$x_ {1} = \frac {41 +\sqrt {381}} {50}\\x_ {2} = \frac {41- \sqrt {381}} {50}$

ANswer:

$x_ {1} = \frac {41 +\sqrt {381}} {50}\\x_ {2} = \frac {41- \sqrt {381}} {50}$

Answer 2

Answer:

$x=-2$

Step-by-step explanation:

The given expression is

$\sqrt{x+3}+4=5$

We group the constant terms on the right hand side to obtain;

$\sqrt{x+3}=5-4$

$\Rightarrow \sqrt{x+3}=1$

We square both sides of the equation to remove the square root.

$\Rightarrow (\sqrt{x+3})^2=1^2$

$x+3=1$

This will simplify to;

$x=1-3$

$x=-2$