Question

$\sqrt{2x - 1} + \sqrt{3x - 2} = \sqrt{4x - 3} + \sqrt{5x - 4}$
$solve$
​

Answer 1

Given,

$\longrightarrow\sqrt{2x-1}+\sqrt{3x-2}=\sqrt{4x-3}+\sqrt{5x-4}$

$\longrightarrow\sqrt{2x-1}-\sqrt{5x-4}=\sqrt{4x-3}-\sqrt{3x-2}$

Squaring both sides,

$\longrightarrow(\sqrt{2x-1}-\sqrt{5x-4})^2=(\sqrt{4x-3}-\sqrt{3x-2})^2$

$\longrightarrow7x-5-2\sqrt{(2x-1)(5x-4)}=7x-5-2\sqrt{(4x-3)(3x-2)}$

$\longrightarrow(2x-1)(5x-4)=(4x-3)(3x-2)$

$\longrightarrow10x^2-13x+4=12x^2-17x+6$

$\longrightarrow2x^2-4x+2=0$

$\longrightarrow x^2-2x+1=0$

$\longrightarrow(x-1)^2=0$

$\longrightarrow\underline{\underline{x=1}}$

Hence solved!

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{Value\:\:of\:\:x\:\:in:\sqrt{2x-1}+\sqrt{3x-2}=\sqrt{4x-3}+\sqrt{5x-4}}$

★═══════════════════════★

♣ ᴀɴꜱᴡᴇʀ :

$\huge\boxed{\bf{x=1}}$

★═══════════════════════★

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\mathrm{Rewrite\:the\:equation\:with\:}\sqrt{2x-1}=u\mathrm{\:and\:}x=\dfrac{u^2+1}{2}$

$\sf{u+\sqrt{3\left(\dfrac{u^2+1}{2}\right)-2}=\sqrt{4\left(\dfrac{u^2+1}{2}\right)-3}+\sqrt{5\left(\dfrac{u^2+1}{2}\right)-4}}$

$\boxed{\bf{u=1}}$

$\mathrm{Substitute\:back}\:u=\sqrt{2x-1},\:\mathrm{solve\:for}\:x$

$\text { Solve } \sqrt{2 x-1}=1$

$\text { Square both sides: }$

$(\sqrt{2x-1})^2=2x-1$

$1^2=1$

$\huge\boxed{\bf{x=1}}$