An equation
\(|\frac{1}{x}-\frac{2}{x+1}|=\frac{2}{x}\)
How many real roots does it have:
a)Exactly 3 real roots
b)Exactly 2 real roots
c)Exactly 1 real root
d)It doesn't have any real roots
| 1/x - 2/(x+1) | = 2/x
Because the left-hand side is contained within absolute value bars, its value must be positive (or zero).
Thus, the right-hand side must also be positive; for 2/x to be positive, x must be positive.
Simplifying the left-hand side: | 1/x - 2/(x + 1) | = | 1(x + 1)/[x(x + 1)] - 2x/[x(x + 1)] |
= | (x + 1 - 2x) / [ x(x + 1) ] |
= | (1 - x) / [ x(x + 1) ] |
Therefore: | (1 - x) / [ x(x + 1) ] | = 2/x
Since it is an absolute value equation, there are two possibilities:
either (1 - x) / [ x(x + 1) ] = 2/x or (1 - x) / [ x(x + 1) ] = -2/x
If (1 - x) / [ x(x + 1) ] = 2/x
---> (1 - x) / [ x(x + 1) ] = 2(x + 1) / [ x(x + 1) ]
---> 1 - x = 2(x + 1)
---> 1 - x = 2x + 2
---> -1 = 3x
---> x = -1/3 <--- But this is impossible, because x must be positive.
If (1 - x) / [ x(x + 1) ] = -2/x
---> (1 - x) / [ x(x + 1) ] = -2(x + 1) / [x(x + 1) ]
---> 1 - x = -2(x + 1)
---> 1 - x = -2x - 2
---> 3 = -x
---> x = -1/3 <--- But this is also impossible.
So, I think that the correct answer is d) it doesn't have any real roots.
Solve for x:
1/x-2/(x+1) = 2/x
Multiply both sides by x:
1-(2 x)/(x+1) = 2
Bring 1-(2 x)/(x+1) together using the common denominator x+1:
(1-x)/(x+1) = 2
Multiply both sides by x+1:
1-x = 2 (x+1)
Expand out terms of the right hand side:
1-x = 2 x+2
Subtract 2 x+1 from both sides:
-3 x = 1
Divide both sides by -3:
Answer: | x = -1/3 Looks like the answer is "C"
Hallo Guest!
An equation
How many real roots does it have:
\(|\frac{1}{x}-\frac{2}{x+1}|=\frac{2}{x}\)
a)Exactly 3 real roots
b)Exactly 2 real roots
c)Exactly 1 real root
d)It doesn't have any real roots
\(|\frac{1}{x}-\frac{2}{x+1}|=\frac{2}{x}\)
\(\frac{x+1-2x}{x²+x}= \frac{2x+2}{x²+x} \) [ * ( x² + x)
x + 1 - 2x = 2x + 2
- 3x = 1
x = - 1 / 3
Sample:
1 / (- 1 / 3) - 2 / (1 - 1 / 3) = 2 / (- 1 / 3)
- 3 - 3 = - 6
- 6 = - 6
Greeting asinus :- )
!
| 1/x - 2/(x+1) | = 2/x
Because the left-hand side is contained within absolute value bars, its value must be positive (or zero).
Thus, the right-hand side must also be positive; for 2/x to be positive, x must be positive.
Simplifying the left-hand side: | 1/x - 2/(x + 1) | = | 1(x + 1)/[x(x + 1)] - 2x/[x(x + 1)] |
= | (x + 1 - 2x) / [ x(x + 1) ] |
= | (1 - x) / [ x(x + 1) ] |
Therefore: | (1 - x) / [ x(x + 1) ] | = 2/x
Since it is an absolute value equation, there are two possibilities:
either (1 - x) / [ x(x + 1) ] = 2/x or (1 - x) / [ x(x + 1) ] = -2/x
If (1 - x) / [ x(x + 1) ] = 2/x
---> (1 - x) / [ x(x + 1) ] = 2(x + 1) / [ x(x + 1) ]
---> 1 - x = 2(x + 1)
---> 1 - x = 2x + 2
---> -1 = 3x
---> x = -1/3 <--- But this is impossible, because x must be positive.
If (1 - x) / [ x(x + 1) ] = -2/x
---> (1 - x) / [ x(x + 1) ] = -2(x + 1) / [x(x + 1) ]
---> 1 - x = -2(x + 1)
---> 1 - x = -2x - 2
---> 3 = -x
---> x = -1/3 <--- But this is also impossible.
So, I think that the correct answer is d) it doesn't have any real roots.
Hallo geno and Alan!
An equation
How many real roots does it have:
\(|\frac{1}{x}-\frac{2}{x+1}|=\frac{2}{x}\)
a)Exactly 3 real roots
b)Exactly 2 real roots
c)Exactly 1 real root
d)It doesn't have any real roots
Of course you are right.
The condition "abs" matters:
Without "absolutely" is x = - 1/3,
with the "absolutely" no root.
It applies
d) It does not have any real roots
Greeting asinus :- )
!
An equation
\(|\frac{1}{x}-\frac{2}{x+1}|=\frac{2}{x} \qquad \qquad x\ne 0\)
How many real roots does it have:
a)Exactly 3 real roots
b)Exactly 2 real roots
c)Exactly 1 real root
d)It doesn't have any real roots
\(\begin{array}{rcll} |\frac{1}{x}-\frac{2}{x+1}| &=& \frac{2}{x} \qquad & | \qquad \text{square both sides} \\ \left(\frac{1}{x}-\frac{2}{x+1} \right)^2 &=& \left(\frac{2}{x} \right)^2 \\ \left(\frac{x+1-2x}{x(x+1)} \right)^2 &=& \frac{4}{x^2} \\ \left(\frac{ 1- x}{x(x+1)} \right)^2 &=& \frac{4}{x^2} \\ \frac{ (1- x)^2}{x^2(x+1)^2} &=& \frac{4}{x^2} \qquad & | \qquad \cdot x^2 \qquad \text{solution }~ x = 0 ~ \text{impossible}\\ \frac{ (1- x)^2}{ (x+1)^2} &=& 4 \\ 4\cdot(x+1)^2 &=& (1- x)^2 \\ 4\cdot( x^2+2x+1) &=& (1-2x+x^2) \\ 4x^2 + 8x+4 &=& 1-2x+x^2 \\ 3x^2 + 10x+3 &=& 0\\ \end{array} \)
\(\begin{array}{|rcll|} \hline 3x^2 + 10x+3 &=& 0 \qquad \qquad a = 3 \qquad b=10 \qquad c = 3 \\\\ x_{1,2} &=& \frac{ -10 \pm \sqrt{10^2-4\cdot 3 \cdot 3} } { 2\cdot 3 } \\\\ x_{1,2} &=& \frac{ -10 \pm \sqrt{100-36} } { 6 } \\\\ x_{1,2} &=& \frac{ -10 \pm \sqrt{64} } {6 } \\\\ x_{1,2} &=& \frac{ -10 \pm 8 } {6 } \\\\ x_{1} &=& \frac{ -10 + 8 } { 6 } \\ x_{1} &=& \frac{ -2 } { 6 } \\ \mathbf{ x_{1} }&\mathbf{=}& \mathbf{ - \frac13 } \qquad &| \qquad \text{solution impossible }~ x > 0 \\\\ x_{2} &=& \frac{ -10 - 8 } { 6 } \\ x_{2} &=& \frac{ -18 } { 6 } \\ \mathbf{ x_{2} }&\mathbf{=}& \mathbf{ - 3 } \qquad &| \qquad \text{solution impossible }~ x > 0 \\\\ \hline \end{array}\)
d) It doesn't have any real roots