Question

Solve the given equation: $27x^{2} -10x+1=0$

Answer 1

$27x^2 - 10x + 1 = 0$

Solve using quadratic formula:

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

Sub a = 27, b = -10, c = 1

$x = \dfrac{10 \pm \sqrt{(-10)^2 - 4(27)(1)} }{2(27)}$

$x = \dfrac{10 \pm \sqrt{100 - 108 }}{54}$

$x = \dfrac{10 \pm \sqrt{-8}}{54}$

$x = \dfrac{10 \pm 2i\sqrt{2}}{54}$

$x = \dfrac{5}{27} \pm \dfrac{i\sqrt{2}}{27}$

$x = \dfrac{5}{27} + \dfrac{i\sqrt{2}}{27} \ or \ x = \dfrac{5}{27} - \dfrac{i\sqrt{2}}{27}$

Answer 2

HEY DEAR ... ✌️

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Here's , Your Answer :-

=) 27x^2 - 10x +1 = 0
find discriminant
D = b^2- 4ac
= (-10)^2 - 4.27.1
= 100 - 108 = -8 < 0
Here D < 0 . it means given polynomial have no real roots .
Hence, we use the concept of complex number.

We know ,
i² = -1 using it here,

27x² -10x +1 = 0

By using quadratic formula ,

x = { -b ± √D}/2a

x = { 10 ± √(-8)}/2.27

= { 10 ± i2√2}/54


= (5 ±i√2)/27

Here, (5 +i√2)/27 , (5 -i√2)/27 are the real roots

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HOPE , IT HELPS ... ✌️