Question

value of x satisfying √x+3+√x-2=5,is
(a) 6
(b) 7
(c) 8
(d) 9

plz help me write all answer​

Answer 1

Answer:

$\sqrt{x + 3} + \sqrt{ \times - 2} = 5$

$put \: \: x = 6$

$\sqrt{6 + 3} + \sqrt{6 - 2}$

$\sqrt{9} + \sqrt{4}$

$3 + 2 = 5$

therefore x=6

Answer 2

${\bigstar \:{\large{\pmb{\sf{\underline{RequirEd \; Solution...}}}}}}$

$\begin{gathered}\\{\small{\underline{\boxed{\pmb{\sf{Equation= \sqrt{x + 3} + \sqrt{x - 2} = 5}}}}}} \\ \\ :\implies {\pmb{\sf{ \sqrt{x + 3 } = 5 - \sqrt{x - 2} }}} \\ \\ \pmb{\sf{ S.O.B.S}}\\ \\ \sf {\sqrt{{(x + 3 )}^{2}} = {(5 - \sqrt{x - 2})}^{2}} \\\\\\ \sf {We \: have \: to \: use \ the \: identity \: given \: below, } \\ \\ {\small{\underline{\boxed{\pmb{\sf{\red{{(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab}}}}}}}\\ \\ \sf Here, \: a \: is \: \pmb{\sf{ 5}}\: and \: b \: is \:\pmb{\sf{ x - 2}}\\ \\ :\implies {\pmb{\sf{x + 3 = {5}^{2} + x - 2 - 10\sqrt{x - 2} }}}\\\\ \sf{By \; cancelling \: x,} \\ \\ :\implies {\pmb{\sf{3 = 25 - 2 - 10\sqrt{ x - 2}}}} \\ \\ :\implies {\pmb{\sf{3 = 23- 10\sqrt{ x - 2}}}} \\ \\ :\implies {\pmb{\sf{10\sqrt{ x - 2} = 23 - 3}}} \\ \\ :\implies {\pmb{\sf{10\sqrt{ x - 2} = 20}}} \\ \\ :\implies {\pmb{\sf{\sqrt{ x - 2} = 2}}} \\ \\ :\implies {\pmb{\sf{x - 2 = 4 }}}\\ \\ :\implies {\pmb{\sf{x = 6}}}\\ \\ \end{gathered}$

Henceforth, we got our solution! Correct option is (A).

$\\ {\bigstar \:{\large{\pmb{\sf{\underline{AdditioNal \; information...}}}}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto \sqrt{\dfrac{a}{b}} \: = \dfrac{\sqrt{a}}{\sqrt{b}}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (a+\sqrt{b}) (a-\sqrt{b}) = a^2 - b}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (\sqrt{a} + \sqrt{b}) (\sqrt{a} - \sqrt{b}) = a - b}}$

$\; \; \; \; \; \; \;{\sf{\leadsto \sqrt{ab} \: = \sqrt{a} \sqrt{b}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (\sqrt{a} + \sqrt{b}) (\sqrt{c} - \sqrt{d}) = \sqrt{ac} + \sqrt{as} + \sqrt{bc} + \sqrt{bd}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (\sqrt{a} + \sqrt{b})^{2} \: = a + 2\sqrt{ab}+b}}$

$\; \; \; \; \; \; \;{\sf{\leadsto ( {a} - {b})^{2} \: = {a}^{2} + 2{ab}+ {b}^{2} }}$