Question

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$\bf\sqrt{2}x^{2}\:-3x+3\sqrt{2}=0$

find x by quadratic equation

Answer 1

Answer :

Comparing the given Equation with ax^2 + bx + c = 0 , we get

a = √2
b = - 3
c = 3√2

Now,

$\sf{b {}^{2} - 4ac = -3 {}^{2} - 4 \times \sqrt{2} \times 3 \sqrt{2} }$

$\sf{= 9 - 24}$

$\sf{ = - 15}$

Now,

Using Formula method,

$\sf{x = \frac{ - b (+ - ) \sqrt{b { }^{2} - 4ac } }{2a}}$

$\sf{x = \frac{ -(- 3) (+ - ) \sqrt{15} }{6}}$

$\sf{x = \frac{ + 1( + - ) \sqrt{15} }{3}}$

Hence,

$\sf{x = \frac{ + 1 + \sqrt{ 15}}{3}}$

OR

$\sf{x = \frac{ + 1 - \sqrt{15} }{3} }$
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Hope it helps..

Answer 2

Solution:

First, We have to Compare the Equation with ax^2 + bx + c = 0 , then we get

a = √2
b = - 3
c = 3√2

Here,

b² - 4ac = -3² - 4 × √2 × 3√2

= 9 - 24

= - 15

Here,

Using Formula,

x = -b (+-)√b²-4ac/2a

x = -(- 3) (+ - )√15/6

x = + 1( + - )√15/3

Hence,

x = +1+√15/3

OR

x = +1-√15/3

Hope it helps..