Question

The number of distinct positive real roots of the equation (x2+6)2−35x2=2x(x2+6) is​

Answer 1

We're given,

$( {x}^{2} + 6) {}^{2} - 35 {x}^{2} = 2x( {x}^{2} + 6)$

Expanding the first term of LHS using (a + b)² = a² + b² + 2ab, with a = x² and b = 6,

$\longrightarrow { ({x}^{2}) }^{2} + {6}^{2} + 2( {x}^{2})(6) - 35 {x}^{2} = 2x( {x}^{2} + 6)$

$\longrightarrow {x}^{4} + 36 + 12 {x}^{2} - 35 {x}^{2} = 2x( {x}^{2} + 6)$

Removing the brackets in RHS,

$\longrightarrow {x}^{4} + 36 + 12 {x}^{2} - 35 {x}^{2} = 2 {x}^{3} + 12x$

$\longrightarrow {x}^{4} - 2 {x}^{3} - 23 {x}^{2} - 12x + 36= 0$

We find $x=1,$ satisfies the equation and is a root of LHS. Hence $(x-1)$ should be a factor.

$\longrightarrow (x - 1) \bigg\{ {x} ^{3} - {x}^{2} - 24x - 36 \bigg\} = 0$

We find $x=-2,$ to be a root of second term. Hence $(x+2)$ should be a factor.

$\longrightarrow (x - 1)(x + 2) \bigg\{ {x}^{2} - 3x - 81 \bigg\} = 0$

$\longrightarrow (x - 1)(x + 2)(x - 6)(x + 3) = 0$

So either (x - 1) = 0, or (x + 2) = 0, or (x - 6) = 0, or (x + 3) = 0.

Hence,

$\longrightarrow \underline{\underline{x = 1,6,-2,-3}}$