Question

(5-x)^1/2=(x+1)

Answer 1

X=1

Answer 2

\((5-x)^\frac{1}{2}=\sqrt{5-x}\) , so :
\(\sqrt{5-x}^2=(x+1)^2\) => \(5-x=x^2+2x+1\), => \(x^2+3x-4=0\)

Answer 3

\[\left( 5-x \right)^{\frac{ 1 }{ 2 }}=x+1\]
squaring both sides
\[5-x=x ^{2}+2x+1,x^2+2x+1+x-5=0,x^2+3x-4=0\]
1*-4=-4
4-1=3
4*-1=-4
write 3x=4x-x
and make factors to find the value of x
Remember 5-x>=0,5>=x or \[x \le 5\]

Answer 4

Solve it as a Quadratic equation

Answer 5

\(\Delta=(3)^2-4(1)(-4)=25\)
\(x1=\frac{-3-5}{2}=\frac{-7}{2}\)

\(x1=\frac{-3+5}{2}=1\)