3x+2/rootx=1,then x-rootx=?
Answers
Answer:
3x+2 =1
To find :
Value of expression :
x - \sqrt{x}x− x
Solution :
The expression given in question is :
\frac{3x + 2}{ \sqrt{x} } = 1
x
3x+2
=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}
x
3x+2
×
x
=1×
x
this will become as,
3x + 2= \sqrt{x}3x+2=
x
(equation 2)
Squaring both sides of equation 2, we get
{(3x + 2)}^{2} = ( {\sqrt{x} } )^{2}(3x+2)
2
=(
x
)
2
so, it becomes,
{(3x + 2)}^{2} = x(3x+2)
2
=x
and
9 {x}^{2} + 12x + 4 = x9x
2
+12x+4=x
so,
9 {x}^{2} + 11x + 4 = 09x
2
+11x+4=0
On solving this equation,
the values of x :
\begin{gathered}(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}\end{gathered}
(1)
x=
18
(−11−
−23)
=
18
(−11−i
23
)
and
(2)
x=
18
(−11+
−23)
=
18
(−11+i
23
)
(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} )−2(
18
(−11−i
23
)
+1)=(
9
−7+i
23
)
The value of given expression :by (3.2)
- 2(\frac{( - 11 + i \sqrt{23}) }{18} + 1) = (\frac{ - 7 - i \sqrt{23} }{9} )−2(
18
(−11+i
23
)
+1)=(
9
−7−i
23
)
So,
Values of x - √x are :
\begin{gathered}\\ \bf (\frac{ - 7 + i \sqrt{23} }{9} ) \\ and \\ \bf (\frac{ - 7 - i \sqrt{23} }{9} )\end{gathered}
(
9
−7+i
23
)
and
(
9
−7−i
23
)