Respuesta :
x = 5
Solution:
Given equation is [tex]x+1=\sqrt{5x+11}[/tex].
[tex]\Rightarrow x+1=\sqrt{5x+11}[/tex]
Squaring on both sides of the equation to remove the square root.
[tex]\Rightarrow (x+1)^2=(\sqrt{5x+11})^2[/tex]
[tex]\Rightarrow (x+1)^2=5x+11[/tex]
Using algebraic identity: [tex](a+b)^2=a^2+2ab+b^2[/tex]
[tex]\Rightarrow x^2+2x(1)+1^2=5x+11[/tex]
[tex]\Rightarrow x^2+2x+1=5x+11[/tex]
Combine all terms in one side of the equation.
[tex]\Rightarrow x^2+2x+1-5x-11=0[/tex]
Arrange like terms together.
[tex]\Rightarrow x^2+2x-5x+1-11=0[/tex]
[tex]\Rightarrow x^2-3x-10=0[/tex]
Now solve by factorization.
[tex]\Rightarrow x^2-5x+2x-10=0[/tex]
[tex]\Rightarrow (x^2-5x)+(2x-10)=0[/tex]
Take common terms on left side of the term.
[tex]\Rightarrow x(x-5)+2(x-5)=0[/tex]
Now, take (x – 5) common on both terms.
[tex]\Rightarrow (x+2)(x-5)=0[/tex]
⇒ x + 2 = 0 (or) x – 5 = 0
⇒ x = –2 (or) x = 5
If we put x = –2 in the given equation,
[tex]-2+1=\sqrt{5(-2)+11}[/tex]
[tex]\Rightarrow-1=1[/tex]
It is false. So, x = –2 is not true.
If we put x = 5 in the given equation,
[tex]5+1=\sqrt{5\times5+11}[/tex]
[tex]5+1=\sqrt{36}[/tex]
[tex]\Rightarrow6=6[/tex]
It is true. So, x = 5 is true.
Hence x = 5 is the solution.
