Question

6+
+[(-5) + (6)] - [4 +61 +
vi. 5+
0
12. Fill each box with a comet integer.
01-
11. - 14+ 0 = 22
N 1-D=9
1 Use r - to make the following statements true.
(-3) 10U (3) (+)
1. (-29) +75 75+(-29)
ii. 30+55 25+ (-60)
in 37+45 D (-35) + (-45)
7++ (-24) (-52) + 36
vi. 75 +98 + (425) -561 +31 + 127
14. The sum of nvo integers is -30. If one of the integers is 15. determine the other.
The difference of an integer a and -13 is 4. Find the value of a.
If the sum of -S and a is-3, then find the additive inverse of a.
submarine was 5827 m below sea level. If it ascends 2200 m,
what is its new position?
ened a bank account by depositing 2500 in her account in May 2012. She deposite
1760 in the month of July 2012. Find her balanc
month of June 2012 and withdrew
TORS
Keya opened a bank account by
1100 in the month of June 201​

Answer 1

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Answer :

The roots of the quadratic equation ,

$\frac{1}{x} - \frac{1}{x - 2} = 3 \: are \: x = \frac{3 + \sqrt[1]{3} }{3} \: and \: x = \frac{3 - \sqrt[1]{3} }{3}$

Explanation :

As given the equation in the form ,

$\frac{1}{x}-\frac{1}{x-2} = 3$

Simplify the above equation

=> (x-2)-x = 3x × (x-2)

=> x-2 - x = 3x² - 6x

=> 3x² - 6x + 2 = 0

As the equation is written in the form ax² + bx + c = 0

$x =\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 3 , b = -6 , c = 2

Put all the values in the above equation

$x =\frac{-(-6)\pm\sqrt{(-6)^{2}-4\times 3\times 2}}{2\times 3}$

$x =\frac{6\pm\sqrt{36-24}}{6}$

$x =\frac{6\pm\sqrt{12}}{6}$

$x =\frac{6\pm2\sqrt{3}}{6}$

$x =\frac{3\pm1\sqrt{3}}{3}$

Thus,

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

$x =\frac{3-1\sqrt{3}}{3}$

Therefore the roots of the quadratic equation $\frac{1}{x}-\frac{1}{x-2} = 3$ are $x =\frac{3+1\sqrt{3}}{3}$,$x =\frac{3-1\sqrt{3}}{3}$

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