+0

0
49
8
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I've got nothing interesting to do right now so I'll just post these. Find the answer or simplify/evaluate the expression.

1. $$x^2 = x-1..How.Many.Real.Solutions?$$
2. $$\sqrt{x+5}-2=6..x= ?$$
3. $$(\sqrt[3]{3}+\sqrt[3]{2})(\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9}).. Easier.than.it.looks$$
4. $$x^4+3x^2+3=1..List.all.possible.solutions$$
5. $$\sqrt{10000}-\sqrt{81}+\sqrt{81}..What.is.the.greatest.possible.value.of.the.expression?$$
6. $$\frac{8}{x}=x+\frac{1}{x}..What.is.the.smallest.possible.value.of.x?$$
7. $$\sqrt{x^3}*\sqrt[3]{x^2}=\frac{\sqrt{y}}{4}..Find.\frac{x}{y}$$

Mathhemathh  Aug 9, 2017
Sort:

#2
+1

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:

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:

No.2:

Solve for x:
sqrt(x + 5) - 2 = 6

sqrt(x + 5) = 8

Raise both sides to the power of two:
x + 5 = 64

Subtract 5 from both sides:

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

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)

x = 1/2 + (i sqrt(3))/2 or x - 1/2 = 1/2 (-i) sqrt(3)

Answer: | x = 1/2 + (i sqrt(3))/2 or x = 1/2 - (i sqrt(3))/2. No "Real Solutions" !!!!

Guest Aug 9, 2017
#3
+4155
+2
 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.     $$\frac{8}{x}=x+\frac1{x} \\~\\ 8=x^2+1 \\~\\ 7=x^2 \\ x=\pm\sqrt{7} \\ \text{ The smallest value is -sqrt(7) .}$$

hectictar  Aug 9, 2017
edited by hectictar  Aug 9, 2017
#4
+797
+2

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.
 $$x^2+1=0$$ $$x^2+2=0$$

 $$x^2=-1$$ $$x^2=-2$$

Take the square root of both sides.
 $$x=\pm\sqrt{-1}$$ $$x=\pm\sqrt{-2}$$

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!

TheXSquaredFactor  Aug 9, 2017
#5
+210
0

Number 5 asks for the greatest solution, and it's not 100.

Mathhemathh  Aug 9, 2017
#6
+75279
+1

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

CPhill  Aug 9, 2017
#7
+6765
0

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...

MaxWong  Aug 11, 2017
#8
+6765
+1

I love math challenges :)

I did math challenges long time ago too... https://web2.0calc.com/questions/i-dont-rly-need-help-some-challenging-questions

MaxWong  Aug 12, 2017

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