Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
1304
8
avatar+351 

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

 

  1. x2=x1..How.Many.Real.Solutions?
  2. x+52=6..x=?
  3. (33+32)(3436+39)..Easier.than.it.looks
  4. x4+3x2+3=1..List.all.possible.solutions
  5. 1000081+81..What.is.the.greatest.possible.value.of.the.expression?
  6. 8x=x+1x..What.is.the.smallest.possible.value.of.x?
  7. x33x2=y4..Find.xy
 Aug 9, 2017
 #2
avatar
+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:
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" !!!!

 Aug 9, 2017
 #3
avatar+9481 
+4
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.            (33+32)(3436+39)=(33)(34)(33)(36)+(33)(39)+(32)(34)(32)(36)+(32)(39)=312318+327+38312+318=327+38=3+2=5

 

5.            1000081+81=10000=104=102=100

 

6.     8x=x+1x 8=x2+1 7=x2x=±7 The smallest value is -sqrt(7) .

 Aug 9, 2017
edited by hectictar  Aug 9, 2017
 #4
avatar+2446 
+2

Let's let everyone jump in! Since hecticlar didn't do #4, I will do number 4.

 

x4+3x2+3=1

 

x4+3x2+3=1 Subtract 1 on both sides.
x4+3x2+2=0 This expression is factorable. Think: What number multiplies to get 2 and add to get 3. That's right! 2 and 1!
(x2+1)(x2+2)=0 Set both factors equal to 0 and solve.
x2+1=0 x2+2=0
 
x2=1 x2=2

 

Take the square root of both sides.
x=±1 x=±2

 

 

 

Now, let's simplify both solutions:

 

x=±1=±i

x=±2=±12=±i2

 

Those are your solutions all done and dusted!
 

 Aug 9, 2017
 #5
avatar+351 
0

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

 Aug 9, 2017
 #6
avatar+130081 
+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

 

 

cool cool cool

 Aug 9, 2017
 #7
avatar+9674 
0

Gonna solve that all :P

1)

x2=x1x2x+1=0Δ=b24ac=(1)24(1)(1)=3No real solutions.

 

2)

x+52=6x+5=8x+5=64x=59

 

3)

Note that:a3+b3=(a+b)(a2ab+b2)(33+32)(3436+39)=(33+32)((32)2(32)(33)+(33)2)=(33)3+(32)3=3+2=5

 

4)

x4+3x2+3=1|u=x2u2+3u+2=0(u+1)(u+2)=0x2+1=0 or x2+2=0x=i,x=i,x=2i,x=2i

 

5) 

1000081+81=10000=100

 

6)

8x=x+1x8=x2+1x27=0x=7 or x=7

 

7) Not enough information...

 Aug 11, 2017
 #8
avatar+9674 
+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

2 Online Users

avatar