Question

if 3x + 2/√x = 1 , then x - √x = ?​

Answer 1

Given :

$\\ \bf \frac{3x + 2}{ \sqrt{x} } = 1$

To find :

Value of expression :

$x - \sqrt{x}$

Solution :

The expression given in question is :

$\frac{3x + 2}{ \sqrt{x} } = 1$

(equation 1)

Multiplying both sides of equation 1 by square root of x,

we get

$\frac{3x + 2}{ \sqrt{x} } \times \sqrt{x} = 1 \times \sqrt{x}$

this will become as,

$3x + 2= \sqrt{x}$

(equation 2)

Squaring both sides of equation 2, we get

${(3x + 2)}^{2} = ( {\sqrt{x} } )^{2}$

so, it becomes,

${(3x + 2)}^{2} = x$

and

$9 {x}^{2} + 12x + 4 = x$

so,

$9 {x}^{2} + 11x + 4 = 0$

On solving this equation,

the values of x :

$(1) \\ x = \frac{( - 11 - \sqrt{ - 23)} }{18} = \frac{( - 11 - i \sqrt{23}) }{18} \\ and \\ (2) \\ x = \frac{( - 11 + \sqrt{ - 23)} }{18} = \frac{( - 11 + i \sqrt{23}) }{18}$

(equation 3)

From equation 2,

3x + 2 = √x

3x - √x = -2

so,

x - √x = -2x - 2 = -2(x + 1)

(equation 4)

So, on putting values of x from equation 3 in equation 4,

The value of given expression :

by (3.1)

$- 2(\frac{( - 11 - i \sqrt{23}) }{18} + 1) = (\frac{ - 7 + i \sqrt{23} }{9} )$

The value of given expression :by (3.2)

$- 2(\frac{( - 11 + i \sqrt{23}) }{18} + 1) = (\frac{ - 7 - i \sqrt{23} }{9} )$

So,

Values of x - √x are :

$\\ \bf (\frac{ - 7 + i \sqrt{23} }{9} ) \\ and \\ \bf (\frac{ - 7 - i \sqrt{23} }{9} )$