How do you factor $5 x^{2} - \frac{5}{2} x + \frac{1}{5} = 0$ $5x^2-5/2x+1/5=0$ ?

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How do you factor $5 x^{2} - \frac{5}{2} x + \frac{1}{5} = 0$ $5x^2-5/2x+1/5=0$ ?

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Answer 1 · 2015-09-28T06:40:46

$x_{1} = \frac{2}{5}$ $x_1 =2/5$
$x_{2} = \frac{1}{10}$ $x_2=1/10$

Explanation:

quadratic formula :
$x_{1},_{2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x_1,_2 =(-b+- sqrt(b^2-4ac))/(2a)$

simplify equation by multiplying both sides with 10
$50 x^{2} - 25 x + 2 = 0$ $50x^2-25x+2=0$

here, a=50, b=-25, c=2

using quadratic formula
$x_{1},_{2} = \frac{- (- 25) \pm \sqrt{{(- 25)}^{2} - 50 \cdot 2}}{2 \cdot 50}$ $x_1,_2=(-(-25)+- sqrt((-25)^2-50*2))/(2*50)$

After solving,
$x_{1} = \frac{25 + \sqrt{225}}{100} = \frac{2}{5}$ $x_1=(25+ sqrt(225))/(100) = 2/5$
$x_{2} = \frac{25 - \sqrt{225}}{100} = \frac{1}{10}$ $x_2=(25- sqrt(225))/(100) = 1/10$

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How do you factor $5 x^{2} - \frac{5}{2} x + \frac{1}{5} = 0$ $5x^2-5/2x+1/5=0$ ?

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