Question

Solve the equation:

$\sf{\sqrt[4]{2x-1}=\dfrac{x^2}{4}+\dfrac{3}{4}}$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: \sqrt[4]{2x - 1} = \dfrac{ {x}^{2} }{4} + \dfrac{3}{4}$

★ Domain of Given equation is

$\rm :\longmapsto\:2x - 1 \geqslant 0 \: \implies \: x \geqslant \dfrac{1}{2}$

★ Let assume that

$\purple{\rm :\longmapsto\:2x - 1 = {y}^{2}} \\ \purple{\rm :\longmapsto\:x=\dfrac{ {y}^{2} + 1}{2} }$

★ On substituting all these values, we get

$\rm :\longmapsto\: \sqrt[4]{ {y}^{2} } = \dfrac{ {\bigg(\dfrac{ {y}^{2} + 1}{2} \bigg) }^{2} }{4} + \dfrac{3}{4}$

$\rm :\longmapsto\: \sqrt{y} = \dfrac{ {y}^{4} + 1 + {2y}^{2} }{16} + \dfrac{3}{4}$

$\rm :\longmapsto\: \sqrt{y} = \dfrac{ {y}^{4} + 1 + {2y}^{2} + 12}{16}$

$\rm :\longmapsto\: \sqrt{y} = \dfrac{ {y}^{4}+ {2y}^{2} + 13}{16}$

★ Again assume that,

$\purple{\rm :\longmapsto\: Let \: \sqrt{y} = z \: \implies \: y = {z}^{2} }$

$\rm :\longmapsto\:z = \dfrac{ {z}^{8} + {2z}^{4} + 13}{16}$

$\rm :\longmapsto\:16z = {z}^{8} + {2z}^{4} + 13$

$\rm :\longmapsto\: {z}^{8} + {2z}^{4} - 16z+ 13 = 0$

Using hit and trial method,

Let z = 1, we get

$\rm :\longmapsto\:1 + 2 - 16 + 13 = 0$

$\rm :\longmapsto\:0 = 0$

$\bf\implies \:z - 1 \: is \: factor.$

So, on long division we get

$\rm :\longmapsto\:(z - 1)( {z}^{7} + {z}^{6} + {z}^{5} + {z}^{4} + {3z}^{3} + {3z}^{2} + 3z - 13) = 0$

$\bf\implies \:z = 1$

Or

$\rm :\longmapsto\:{z}^{7} + {z}^{6} + {z}^{5} + {z}^{4} + {3z}^{3} + {3z}^{2} + 3z - 13 = 0$

Again, using hit and trial method,

Let z = 1, we get

$\rm :\longmapsto\:1 + 1 + 1 + 1 + 3 + 3 + 3 - 13 = 0$

$\rm :\longmapsto\:0 = 0$

$\bf\implies \:z - 1 \: is \: a \: factor.$

So, on long division we get

$\rm :\longmapsto\:(z - 1)( {z}^{6} + {2z}^{5} + {3z}^{4} + {4z}^{3} + {7z}^{2} + 10z + 13) = 0$

$\bf\implies \:z = 1$

or

$\rm :\longmapsto\:{z}^{6} + {2z}^{5} + {3z}^{4} + {4z}^{3} + {7z}^{2} + 10z + 13= 0$

which can never be zero for any positive real value of z.

So,

Hence, we have

$\rm :\longmapsto\:z = 1$

$\rm :\longmapsto\: \sqrt{y} = 1$

$\rm :\implies\:y = 1$

$\rm :\longmapsto\: \sqrt{2x - 1} = 1$

$\rm :\longmapsto\:2x - 1 = 1$

$\rm :\longmapsto\:2x = 1 + 1$

$\rm :\longmapsto\:2x =2$

$\bf\implies \:x = 1$