Answer:
[tex]x=-2[/tex]
Step-by-step explanation:
The given expression is
[tex]\sqrt{x+3}+4=5[/tex]
We group the constant terms on the right hand side to obtain;
[tex]\sqrt{x+3}=5-4[/tex]
[tex]\Rightarrow \sqrt{x+3}=1[/tex]
We square both sides of the equation to remove the square root.
[tex]\Rightarrow (\sqrt{x+3})^2=1^2[/tex]
[tex]x+3=1[/tex]
This will simplify to;
[tex]x=1-3[/tex]
[tex]x=-2[/tex]
For this case we must solve the following equation:
[tex]\sqrt {x+3}+4 = 5x[/tex]
We subtract 4 on both sides of the equation:
[tex]\sqrt {x+3}+4-4 = 5x-4\\\sqrt {x+3} = 5x-4[/tex]
We square both sides of the equation:
[tex](\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\\[/tex]
We have an equation of the form:
[tex]ax ^ 2+bx+c = 0[/tex]
Where:
[tex]a = 25\\b = -41\\c = 13[/tex]
The solutions are given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}[/tex]
Substituting:
[tex]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}[/tex]
So, the roots are:
[tex]x_ {1} = \frac {41 +\sqrt {381}} {50}\\x_ {2} = \frac {41- \sqrt {381}} {50}[/tex]
ANswer:
[tex]x_ {1} = \frac {41 +\sqrt {381}} {50}\\x_ {2} = \frac {41- \sqrt {381}} {50}[/tex]
