Question

the roots of 100x^2-20x+1=0are???​

Answer 1

hope it help you

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Answer 2

Answer:

$100x^{2} - 20x + 1 = 0$

a = 100

b = - 20

c = 1

using sridharacharya formula

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

putting the values of a , b & c

$x = \frac {- (-20) ± \sqrt{(-20)^{2} - 4(100)(1)}}{2(100)}\\\\x = \frac {20 ± \sqrt{400 - 400}}{200}\\\\x = \frac {20 ± \sqrt{0}}{200}\\\\x = \frac {20 ± 0}{200}$

therefore

$x = \frac {20 - 0}{200}\\\\x = \frac{20}{200}\\\\x = \frac{1}{10}$

&

$x = \frac {20 + 0}{200}\\\\x = \frac{20}{200}\\\\x = \frac{1}{10}$

therefore the roots of the equation is $\frac{1}{10}$ & $\frac{1}{10}$