solve the equation √3x²-√2x+3√3=0
Answers
Given : \sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0Given:
3
x
2
−2
2
x−2
3
=0
On comparing with ax^2 + bx + c = 0, we get
= > a = \sqrt{3},b = -2\sqrt{2},c = -2\sqrt{3}=>a=
3
,b=−2
2
,c=−2
3
(1)
x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}x=
2a
−b+
b
2
−4ac
= > \frac{-(-2\sqrt{2})+\sqrt{(-2\sqrt{2})^2 - 4\sqrt{3}(-2\sqrt{3})}}{2\sqrt{3}}=>
2
3
−(−2
2
)+
(−2
2
)
2
−4
3
(−2
3
)
= > \frac{2\sqrt{2}\sqrt{(2\sqrt{2})^2 + 4\sqrt{3} * 2\sqrt{3}}}{2\sqrt{3}}=>
2
3
2
2
(2
2
)
2
+4
3
∗2
3
= > \frac{2\sqrt{2}+\sqrt{(2\sqrt{2})^2 + 24}}{2\sqrt{3}}=>
2
3
2
2
+
(2
2
)
2
+24
= > \frac{2\sqrt{2} + \sqrt{32}}{2\sqrt{3}}=>
2
3
2
2
+
32
= > \frac{2\sqrt{2} + 4\sqrt{2}}{2\sqrt{3}}=>
2
3
2
2
+4
2
= > \frac{6\sqrt{2}}{2\sqrt{3}}=>
2
3
6
2
= > \frac{3\sqrt{2}}{\sqrt{3}}=>
3
3
2
= > \sqrt{6}=>
6
---------------------------------------------------------------------------------------------------------------
(2)
= > x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}=>x=
2a
−b−
b
2
−4ac
= > x = \frac{-(-2\sqrt{2})-\sqrt{(2\sqrt{2})^2+4\sqrt{3}*2\sqrt{3}}}{2\sqrt{3}}=>x=
2
3
−(−2
2
)−
(2
2
)
2
+4
3
∗2
3
= >\frac{2\sqrt{2}-{4\sqrt{2}}}{2\sqrt{3}}=>
2
3
2
2
−4
2
= > \frac{-2\sqrt{2}}{2\sqrt{3}}=>
2
3
−2
2
= > -\sqrt{\frac{2}{3}}=>−
3
2
---------------------------------------------------------------------------------------------------------------
Therefore the required quadratic solutions are|:
= > x = \boxed {\sqrt{6}, -\sqrt{\frac{2}{3}} }=>x=
6
,−
3
2
Hope this helps
Answer:
x ={-b+-√(b²-4ac)}/2a
x=(√2+-6)/6
x1=√2/6+1. X2=√2/6-1