I've got nothing interesting to do right now so I'll just post these. Find the answer or simplify/evaluate the expression.
No. 6:
Solve for x over the real numbers:
8/x = x + 1/x
Bring x + 1/x together using the common denominator x:
8/x = (x^2 + 1)/x
Multiply both sides by x:
8 = x^2 + 1
8 = x^2 + 1 is equivalent to x^2 + 1 = 8:
x^2 + 1 = 8
Subtract 1 from both sides:
x^2 = 7
Take the square root of both sides:
Answer: | x = sqrt(7) or x = -sqrt(7)
No.7:
x/y = 1/(16 (x^2)^(5/3))
No.5:
Simplify the following:
sqrt(10000) - sqrt(81) + sqrt(81)
sqrt(10000) - sqrt(81) + sqrt(81) = 100:
Answer: | 100
No.4:
Solve for x:
x^4 + 3 x^2 + 3 = 1
Subtract 1 from both sides:
x^4 + 3 x^2 + 2 = 0
Substitute y = x^2:
y^2 + 3 y + 2 = 0
The left hand side factors into a product with two terms:
(y + 1) (y + 2) = 0
Split into two equations:
y + 1 = 0 or y + 2 = 0
Subtract 1 from both sides:
y = -1 or y + 2 = 0
Substitute back for y = x^2:
x^2 = -1 or y + 2 = 0
Take the square root of both sides:
x = i or x = -i or y + 2 = 0
Subtract 2 from both sides:
x = i or x = -i or y = -2
Substitute back for y = x^2:
x = i or x = -i or x^2 = -2
Take the square root of both sides:
Answer: | x = i or x = -i or x = i sqrt(2) or x= -i sqrt(2)
No.3:
Expand the following:
(3^(1/3) + 2^(1/3)) (4^(1/3) - 6^(1/3) + 9^(1/3))
4^(1/3) = (2^2)^(1/3):
(3^(1/3) + 2^(1/3)) (2^(2/3) - 6^(1/3) + 9^(1/3))
9^(1/3) = (3^2)^(1/3):
(3^(1/3) + 2^(1/3)) (2^(2/3) - 6^(1/3) + 3^(2/3))
(2^(2/3) + 3^(2/3) - 6^(1/3)) (2^(1/3) + 3^(1/3)) = 2^(1/3) (2^(2/3) + 3^(2/3) - 6^(1/3)) + 3^(1/3) (2^(2/3) + 3^(2/3) - 6^(1/3)):
2^(1/3) (2^(2/3) + 3^(2/3) - 6^(1/3)) + 3^(1/3) (2^(2/3) + 3^(2/3) - 6^(1/3))
2^(1/3) (2^(2/3) + 3^(2/3) - 6^(1/3)) = 2^(1/3)×2^(2/3) + 2^(1/3)×3^(2/3) + 2^(1/3) (-6^(1/3)):
2^(1/3)×2^(2/3) + 2^(1/3)×3^(2/3) - 2^(1/3) 6^(1/3) + 3^(1/3) (2^(2/3) + 3^(2/3) - 6^(1/3))
2^(1/3)×2^(2/3) = 2^(1/3 + 2/3):
2^(1/3 + 2/3) + 2^(1/3)×3^(2/3) - 2^(1/3) 6^(1/3) + 3^(1/3) (2^(2/3) + 3^(2/3) - 6^(1/3))
2^(1/3 + 2/3) = 2:
2 + 2^(1/3)×3^(2/3) - 2^(1/3) 6^(1/3) + 3^(1/3) (2^(2/3) + 3^(2/3) - 6^(1/3))
2^(1/3) 6^(1/3) = (2×6)^(1/3):
2 + 2^(1/3)×3^(2/3) - (2×6)^(1/3) + 3^(1/3) (2^(2/3) + 3^(2/3) - 6^(1/3))
2×6 = 12:
2 + 2^(1/3)×3^(2/3) - (12 )^(1/3) + 3^(1/3) (2^(2/3) + 3^(2/3) - 6^(1/3))
12^(1/3) = (2^2×3)^(1/3) = 2^(2/3) 3^(1/3):
2 + 2^(1/3)×3^(2/3) - 2^(2/3) 3^(1/3) + 3^(1/3) (2^(2/3) + 3^(2/3) - 6^(1/3))
3^(1/3) (2^(2/3) + 3^(2/3) - 6^(1/3)) = 3^(1/3)×2^(2/3) + 3^(1/3)×3^(2/3) + 3^(1/3) (-6^(1/3)):
2 + 2^(1/3)×3^(2/3) - 2^(2/3) 3^(1/3) + 3^(1/3)×2^(2/3) + 3^(1/3)×3^(2/3) - 3^(1/3) 6^(1/3)
3^(1/3)×3^(2/3) = 3^(1/3 + 2/3):
2 + 2^(1/3)×3^(2/3) - 2^(2/3) 3^(1/3) + 3^(1/3)×2^(2/3) + 3^(1/3 + 2/3) - 3^(1/3) 6^(1/3)
3^(1/3 + 2/3) = 3:
2 + 2^(1/3)×3^(2/3) - 2^(2/3) 3^(1/3) + 3^(1/3)×2^(2/3) + 3 - 3^(1/3) 6^(1/3)
3^(1/3) 6^(1/3) = (3×6)^(1/3):
2 + 2^(1/3)×3^(2/3) - 2^(2/3) 3^(1/3) + 3^(1/3)×2^(2/3) + 3 - (3×6)^(1/3)
3×6 = 18:
2 + 2^(1/3)×3^(2/3) - 2^(2/3) 3^(1/3) + 3^(1/3)×2^(2/3) + 3 - (18 )^(1/3)
18^(1/3) = (2×3^2)^(1/3) = 2^(1/3)×3^(2/3):
2 + 2^(1/3)×3^(2/3) - 2^(2/3) 3^(1/3) + 3^(1/3)×2^(2/3) + 3 - 2^(1/3)×3^(2/3)
2 + 2^(1/3)×3^(2/3) - 2^(2/3) 3^(1/3) + 3^(1/3)×2^(2/3) + 3 - 2^(1/3)×3^(2/3) = 5:
Answer: | 5
No.2:
Solve for x:
sqrt(x + 5) - 2 = 6
Add 2 to both sides:
sqrt(x + 5) = 8
Raise both sides to the power of two:
x + 5 = 64
Subtract 5 from both sides:
Answer: | x = 59
No.1:
Solve for x:
x^2 = x - 1
Subtract x - 1 from both sides:
x^2 - x + 1 = 0
Subtract 1 from both sides:
x^2 - x = -1
Add 1/4 to both sides:
x^2 - x + 1/4 = -3/4
Write the left hand side as a square:
(x - 1/2)^2 = -3/4
Take the square root of both sides:
x - 1/2 = (i sqrt(3))/2 or x - 1/2 = 1/2 (-i) sqrt(3)
Add 1/2 to both sides:
x = 1/2 + (i sqrt(3))/2 or x - 1/2 = 1/2 (-i) sqrt(3)
Add 1/2 to both sides:
Answer: | x = 1/2 + (i sqrt(3))/2 or x = 1/2 - (i sqrt(3))/2. No "Real Solutions" !!!!
1. | x2 = x - 1 | ||
x2 - x + 1 = 0 | |||
x = ( 1 ± √[ 1 - 4] )/ 2 | |||
x = ( 1 ± √3 i ) / 2 | There are no real solutions for x . | ||
2. | √[ x + 5] - 2 = 6 | ||
√[ x + 5] = 8 | |||
x + 5 = 82 | |||
x = 82 - 5 | |||
x = 59 |
3. \(\, \ \, \ \, (\sqrt[3]{3}+\sqrt[3]{2})(\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9}) \\ =(\sqrt[3]3)(\sqrt[3]{4})-(\sqrt[3]3)(\sqrt[3]{6})+(\sqrt[3]3)(\sqrt[3]{9}) + (\sqrt[3]{2})(\sqrt[3]{4})-(\sqrt[3]{2})(\sqrt[3]{6})+(\sqrt[3]{2})(\sqrt[3]{9}) \\ =\sqrt[3]{12}-\sqrt[3]{18}+\sqrt[3]{27} + \sqrt[3]{8}-\sqrt[3]{12}+\sqrt[3]{18} \\ =\sqrt[3]{27} + \sqrt[3]{8} \\ =3+2\\ =5\)
5. \(\, \ \, \ \, \sqrt{10000}-\sqrt{81}+\sqrt{81} \\ =\sqrt{10000} \\ =\sqrt{10^4} \\ =10^2 \\ =100\)
6. \(\frac{8}{x}=x+\frac1{x} \\~\\ 8=x^2+1 \\~\\ 7=x^2 \\ x=\pm\sqrt{7} \\ \text{ The smallest value is -sqrt(7) .}\)
Let's let everyone jump in! Since hecticlar didn't do #4, I will do number 4.
\(x^4+3x^2+3=1\)
\(x^4+3x^2+3=1\) | Subtract 1 on both sides. | ||
\(x^4+3x^2+2=0\) | This expression is factorable. Think: What number multiplies to get 2 and add to get 3. That's right! 2 and 1! | ||
\((x^2+1)(x^2+2)=0\) | Set both factors equal to 0 and solve. | ||
| |||
| Take the square root of both sides. | ||
|
Now, let's simplify both solutions:
\(x=\pm\sqrt{-1}=\pm i\)
\(x=\pm\sqrt{-2}=\pm\sqrt{-1*2}=\pm i\sqrt{2}\)
Those are your solutions all done and dusted!
Note that for (1) we can rearrange it as
x^2 - x + 1 = 0
The discriminant for this is b^2 - 4ac = (-1)^2 - 4 (1) (1) = -3
And when the discriminant < 0, we will have no real solutions
Gonna solve that all :P
1)
\(x^2=x-1\\ x^2-x+1=0\\ \Delta = b^2 - 4ac = (-1)^2-4(1)(1)=-3\\ \therefore\text{No real solutions.}\)
2)
\(\sqrt{x+5}-2=6\\ \sqrt{x+5}=8\\ x+5=64\\ x=59\)
3)
\(\text{Note that:}\\\boxed{a^3+b^3=(a+b)(a^2-ab+b^2)}\\ (\sqrt[3]{3}+\sqrt[3]{2})(\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9})\\ =(\sqrt[3]{3}+\sqrt[3]{2})\left((\sqrt[3]{2})^2-(\sqrt[3]{2})(\sqrt[3]{3})+(\sqrt[3]{3})^2\right)\\ =(\sqrt[3]{3})^3+(\sqrt[3]{2})^3\\ =3+2\\ =5\)
4)
\(x^4+3x^2+3=1\; |u=x^2\\ u^2+3u+2=0\\ (u+1)(u+2)=0\\ x^2+1=0\text{ or }x^2+2=0\\ x=i,x=-i,x=\sqrt{2}i,x=-\sqrt{2}i\)
5)
\(\sqrt{10000}-\sqrt{81}+\sqrt{81}\\ =\sqrt{10000}\\ =100\)
6)
\(\dfrac{8}{x}=x+\dfrac{1}{x}\\ 8=x^2+1\\ x^2-7=0\\ x=\sqrt7\text{ or }x=-\sqrt7\)
7) Not enough information...