x=3 , 8
Explanationremember the square of a binomyal
[tex](a\pm b)^2=a^2\pm2ab+b^2[/tex]
Step 1
given
[tex]\sqrt{x+1}\text{ =x-5}[/tex]
we need to isolate x, so
a) rise each side to power 2
[tex]\begin{gathered} \sqrt{x+1}\text{ =x-5} \\ (\sqrt{x+1})\text{ }=(x-5)^2 \\ x+1=x^2-2*5*x+5^2 \\ x+1=x^2-10x+25 \\ subtrac\text{ x in both sides} \\ x+1-x=x^2-10x+25-x \\ 1=x^2-11x+25 \\ subtract\text{ 1 in both sides} \\ 1-1=x^2-11x+25-1 \\ hence \\ x^2-11x+24=0 \end{gathered}[/tex]
Step 2
solve the quadratic equation:
b) use the quadratic formula
[tex]\begin{gathered} it\text{ says} \\ for\text{ ax}^2+bx+c=0 \\ the\text{ solution for x is} \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{gathered}[/tex]
so
i)let
[tex]\begin{gathered} ax^2+bx+c=x^2-11x+24 \\ so \\ a=1 \\ b=-11 \\ c=24 \end{gathered}[/tex]
ii) now, replace in the formula
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x=\frac{-(-11)\pm\sqrt{-11^2-4(1)(24)}}{2(1)} \\ x=\frac{11\pm\sqrt{121-96}}{2} \\ x=\frac{11\pm\sqrt{25}}{2}=\frac{11\pm5}{2} \\ so \\ x_1=\frac{11+5}{2}=\frac{16}{2}=8 \\ x_2=\frac{11-5}{2}=\frac{6}{2}=3 \end{gathered}[/tex]
therefore, the solutions are x= 3 and x= 8
so, the answer is
x=3 , 8
I hope this helps you
