Question

Solve the equation:


`4 \sqrt{3 - (1)/(x) } - \sqrt{ (x)/(3x - 1) } = 3`
Are the solutions of the equation solutions of

`\frac{(1 - 2x)^(2) * (x^(2) - 9) }{ -{x}^(2) - x + 6} \geqslant 0`
? ​

Answer 1

Explanation:

`4\sqrt{3-(1)/(x) } -\sqrt{(x)/(3x-1) } =3\\4\sqrt{(3x-1)/(x) } -\sqrt{(x)/(3x-1) } =3\\\\put ~\sqrt{(3x-1)/(x)} =a\\4a-(1)/(a) =3\\4a^2-1-3a=0\\4a^2-4a+1a-1=0\\4a(a-1)+1(a-1)=0\\(a-1)(4a+1)=0\\a=1,-(1)/(4)\\`

`\sqrt{(3x-1)/(x) } =1\\3x-1=x\\3x-x=1\\2x=1\\x=(1)/(2)`

when a=-1/4

`\sqrt{(3x-1)/(x) } =-(1)/(4) \\`

`(3x-1)/(x) =(1)/(16) \\48x-16=x\\47x=16\\x=(16)/(47)`

`((1-2x)^2 * (x^2-9))/(-x^2-x+6) \geq 0,\\both ~numerator ~and~denominator ~are~of~same~sign.\\((1-2x)^2(x^2-9))/(-x^2-3x+2x+6) \geq 0\\((1-2x)^2(x+3)(x-3))/(-x(x+3)+2(x+3)) \geq 0\\((1-2x)^2(x+3)(x-3))/((x+3)(-x+2)) \geq 0\\(1-2x)^2\geq 0~(always)\\x\\eq -3\\case.~1.\\both~numerator~and~denominator >0\\(x-3)/(-x+2) \geq 0\\x-3\geq 0\\x\geq 3\\-x+2> 0\\-x>-2\\x<2\\so~ solution~ is~ (-\infty,-3)U(-3,2)U[3,\infty)`

case 2.

both numerator and denominator are negative.

`x-3\leq 0\\x\leq 3\\-x+2\leq 0\\-x\leq -2\\x\geq 2\\solution~is~(-\infty,-3)U(-3,3]U[2,\infty)`