+118687 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.
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
Add 6 to both sides:
x = 6 or 3 x - 2 = 0
Add 2 to both sides:
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:
Answer: | x = 6
+118687 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.
