Question

${\boxed {\LARGE{\mathfrak {solve\:it:-}}}}$
$\tt \sqrt{2x+9}+x=13$​

Answer 1

$\tt \sqrt{2x+9}+x=13 \\ {\tiny{ \text{transposing \: x \: to \: RHS} }} \\ \implies \tt \sqrt{2x + 9} = 13 - x \\ \tiny{ \text{squaring \: both \: sides}} \\ \implies \tt2x + 9 = 169 + {x}^{2} - 26x \\ \tiny{ \text{transposing \: RHS \: to \: LHS}}\\ \implies \tt0 = {x}^{2} - 28x + 160 \\ \tiny{ \text{switching \: the \: LHS \: \& \: RHS}} \\ \implies \tt {x}^{2} - 28x + 160 = 0 \\ \tiny{ \downarrow\text{solving \: the \: equation }}\downarrow \\ \implies \tt {x}^{2} - 20x - 8x + 160 = 0 \\ \implies \tt x(x - 20) - 8(x - 20) = 0 \\ \implies \tt(x - 20)(x - 8) = 0 \\ \tiny{ \text{equating \: each \: factors \: with \: zero}} \\ \\ { \text{we \: obtained,}} \\ \sf \: \: \: x = \red{20} \: \: \: or \: \: \: \red8$