If, 2,5,8=122348, 3,5,4=181224, 7,4,6=352230, 5,3,10 =203340 then 3,9,4=?
Answers
Answer:
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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}x1−x−21=3arex=33+13andx=33−13
Explanation :
As given the equation in the form ,
\frac{1}{x}-\frac{1}{x-2} = 3x1−x−21=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}x=2a−b±b2−4ac
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=2×3−(−6)±(−6)2−4×3×2
x =\frac{6\pm\sqrt{36-24}}{6}x=66±36−24
x =\frac{6\pm\sqrt{12}}{6}x=66±12
x =\frac{6\pm2\sqrt{3}}{6}x=66±23
x =\frac{3\pm1\sqrt{3}}{3}x=33±13
Thus,
x =\frac{3+1\sqrt{3}}{3}x=33+13
x =\frac{3-1\sqrt{3}}{3}x=33−13
Therefore the roots of the quadratic equation \frac{1}{x}-\frac{1}{x-2} = 3x1−x−21=3 are x =\frac{3+1\sqrt{3}}{3}x=33+13 ,x =\frac{3-1\sqrt{3}}{3}x=33−13
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Answer:
Explanation:
2,5,8=122348and3,5,4=181224and7,4,6=352230and5,3,10=203340and3,9,4