+0

# Solve fo x : ²-

+1
162
4
+302

Solve fo x : $$\sqrt{}3x$$²-$$2\sqrt{}2x-2\sqrt{}3=0$$

SARAHann  Apr 1, 2017

#3
+90508
+4

Hi Sara :)

Lets see :)

$$\sqrt3\;x^2-2\sqrt2\;x-2\sqrt3=0\\~\\ \triangle=(-2\sqrt2)^2-4*\sqrt3* -2\sqrt3 \\ \triangle=(8+24) \\ \triangle=32\qquad \text{Positive so this means it has 2 real roots} \\~\\$$

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-b \pm \sqrt{\triangle} \over 2a}\\ x = {2\sqrt2 \pm \sqrt{32} \over 2\sqrt3}\\ x = {2\sqrt2 \pm 4\sqrt{2} \over 2\sqrt3}\\ \\~\\ x = {2\sqrt2 \pm 4\sqrt{2} \over 2\sqrt3}\times \frac{\sqrt3}{\sqrt3}\\ x = {2\sqrt6 \pm 4\sqrt{6} \over 6}\\ x = {\sqrt6 \pm 2\sqrt{6} \over 3}\\ x = {\sqrt6 (1\pm2) \over 3}\\ x=\frac{3\sqrt6}{3}\qquad or \qquad x=\frac{-\sqrt6}{3}\\ x=\sqrt6\qquad or \qquad x=\frac{-\sqrt6}{3}\\$$

I checked this answer by graphing it and it is correct.

Melody  Apr 1, 2017
Sort:

#1
+1

Solve for x: P.S. This is how I read it: sqrt(3x^2) - 2sqrt(2x) - 2sqrt(3) = 0, solve for x
-2 sqrt(3) - 2 sqrt(2) sqrt(x) + sqrt(3) sqrt(x^2) = 0

Add 2 sqrt(3) to both sides:
sqrt(3) x - 2 sqrt(2) sqrt(x) = 2 sqrt(3)

Subtract sqrt(3) x from both sides:
-2 sqrt(2) sqrt(x) = 2 sqrt(3) - sqrt(3) x

Raise both sides to the power of two:
8 x = (2 sqrt(3) - sqrt(3) x)^2

Expand out terms of the right hand side:
8 x = 3 x^2 - 12 x + 12

Subtract 3 x^2 - 12 x + 12 from both sides:
-3 x^2 + 20 x - 12 = 0

The left hand side factors into a product with three terms:
-(x - 6) (3 x - 2) = 0

Multiply both sides by -1:
(x - 6) (3 x - 2) = 0

Split into two equations:
x - 6 = 0 or 3 x - 2 = 0

x = 6 or 3 x - 2 = 0

x = 6 or 3 x = 2

Divide both sides by 3:
x = 6 or x = 2/3

-2 sqrt(3) - 2 sqrt(2) sqrt(x) + sqrt(3) sqrt(x^2) ⇒ -2 sqrt(3) - 2 sqrt(2) sqrt(2/3) + sqrt(3) sqrt((2/3)^2) = -8/sqrt(3) ≈ -4.6188:
So this solution is incorrect

-2 sqrt(3) - 2 sqrt(2) sqrt(x) + sqrt(3) sqrt(x^2) ⇒ -2 sqrt(3) - 2 sqrt(2) sqrt(6) + sqrt(3) sqrt(6^2) = 0:
So this solution is correct

The solution is:

Guest Apr 1, 2017
#2
+302
+1

could u explain using latex? i did'nt get it

SARAHann  Apr 1, 2017
#3
+90508
+4

Hi Sara :)

Lets see :)

$$\sqrt3\;x^2-2\sqrt2\;x-2\sqrt3=0\\~\\ \triangle=(-2\sqrt2)^2-4*\sqrt3* -2\sqrt3 \\ \triangle=(8+24) \\ \triangle=32\qquad \text{Positive so this means it has 2 real roots} \\~\\$$

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-b \pm \sqrt{\triangle} \over 2a}\\ x = {2\sqrt2 \pm \sqrt{32} \over 2\sqrt3}\\ x = {2\sqrt2 \pm 4\sqrt{2} \over 2\sqrt3}\\ \\~\\ x = {2\sqrt2 \pm 4\sqrt{2} \over 2\sqrt3}\times \frac{\sqrt3}{\sqrt3}\\ x = {2\sqrt6 \pm 4\sqrt{6} \over 6}\\ x = {\sqrt6 \pm 2\sqrt{6} \over 3}\\ x = {\sqrt6 (1\pm2) \over 3}\\ x=\frac{3\sqrt6}{3}\qquad or \qquad x=\frac{-\sqrt6}{3}\\ x=\sqrt6\qquad or \qquad x=\frac{-\sqrt6}{3}\\$$

I checked this answer by graphing it and it is correct.

Melody  Apr 1, 2017
#4
+302
0

Thx Mel!

SARAHann  Apr 3, 2017

### 20 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details