Question

Solve:
$\sqrt{2x - 1} - \sqrt{x - 4} = 2$
​

Answer 1

Question :-

Solve it :

$\mapsto \sf\sqrt{2x - 1} - \sqrt{x - 4} = 2 \:$

Answer :-

$\to \boxed{ \sf x = 5 \: ,13 \: } \:$

Solution :-

We have ,

$\to \sf\sqrt{2x - 1} - \sqrt{x - 4} = 2 \:$

We can write it :

$\to \sf \: \sqrt{2x - 1} = 2 + \sqrt{x - 4} \\ \\ \sf \red{ squaring \: on \: both \: sides \: } \\ \\ \to \sf {( \sqrt{2x - 1} )}^{2} = {(2 + \sqrt{x - 4}) }^{2} \\ \\ \green{ \boxed{ \red{ \because }\sf {(a + b)}^{2} = \orange{ {a}^{2} + {b}^{2} }+ 2ab}} \\ \\ \to \sf2x - 1 = {2}^{2} + {( \sqrt{x - 4} )}^{2} + 2 \times 2 \sqrt{x - 4} \\ \\ \to \sf \: 2x - 1 = 4 + x - 4 + 4 \sqrt{x - 4} \\ \\ \to \sf \: 2x - 1 - x = 4 \sqrt{x - 4} \\ \\ \to \sf \: x - 1 = 4 \sqrt{x - 4} \\ \\ \sf \red{ again \: squaring} \: \green{on \: both \: sides \:} \\ \\ \to \sf {(x - 1)}^{2} = {(4 \sqrt{x - 4} )}^{2} \\ \\ \to \sf {x}^{2} + 1 - 2x = 16(x - 4) \\ \\ \to \sf {x}^{2} - 2x + 1 - 16x + 64 = 0 \\ \\ \to \sf {x}^{2} - 18x + 65 = 0$

Split the middle term :

$\to \sf {x}^{2} - 13x - 5x + 65 = 0 \\ \\ \to \sf \: x(x - 13) - 5(x - 13) = 0 \\ \\ \to \sf \: (x - 13)(x - 5) = 0 \\ \\ \star \bf \underline{ \: case \: (1) }\: if \: \\ \\ \to \sf \: x - 13 = 0 \\ \\ \to \boxed{ \sf x = 13} \\ \\ \star \: \bf \underline{case \: (2)} \: if \: \\ \\ \to \sf x - 5 = 0 \\ \\ \to \boxed{ \sf x = 5}$

hence x = 5 ,13