Solve the equation (x^3 - x)^1/2 + (2x - 1)^1/2 = (x^3 + x - 1)1/2.

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To solve the equation and to eliminate the square roots, we'll square raise both sides:

[sqrt(x^3 - x) + sqrt(2x - 1)]^2= [sqrt(x^3 + x - 1)]^2

x^3 - x + 2x - 1 + 2sqrt[(x^3 - x)(2x - 1)] = x^3 + x - 1

We'll combine and eliminate like terms and we'll get:

2sqrt[(x^3 - x)(2x - 1)] = 0

We'll divide by 2:

[(x^3 - x)(2x - 1)] = 0

We'll set each factor as zero:

x^3 - x = 0

We'll factorize:

x(x^2-1) = 0

We'll expand the difference of squares:

x(x-1)(x+1) = 0

We'll set each factor as zero:

x1 = 0

x-1 = 0

x2 = 1

x+1 = 0

x3 = -1

2x - 1 = 0

2x = 1

x4 = 1/2

Now, we'll check the found results in equation:

For x1 = 0

sqrt(x^3 - x) + sqrt(2x - 1)= sqrt(x^3 + x - 1)

sqrt(0) + sqrt( - 1)= sqrt( - 1)

But sqrt -1 is impossible!

For x2 = 1

sqrt(1 -1) + sqrt(2 - 1)= sqrt(1 + 1- 1)

0 + 1 = 1

1=1

So x = 1 is admissible.

For x3 = -1

sqrt(x^3 - x) + sqrt(2x - 1)= sqrt(x^3 + x - 1)

sqrt(-1 + 1) + sqrt(-2 - 1)= sqrt(-1 -1- 1)

sqrt -3 = sqrt-3

But, sqrt-3 is impossible!

For x = 1/2

sqrt(x^3 - x) + sqrt(2x - 1)= sqrt(x^3 + x - 1)

sqrt(1/8- 1/2) + sqrt(2/2 - 1)= sqrt(1/8 +1/2 - 1)

But sqrt -3/8 is impossible!

So, the only admissible solution of the equation is x = 1.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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(x^3-x)^1/2 +(2x-1)^(1/2) = (x^3+x-1)^(1/2).

Solution:

Rewrite

(2x-1)^1/2 = (x^3+x-1)^(1/2)-(x^3-x)^(1/2).....(1) Rationalise right side numerator.

(2x-1)^(1/2) =  {x^3+x-1 -(x^3-x)}/ {x^3+x-1) +(x^3 -x)^(1/2)}

(2x-1)^(1/2) =  (2x-1) / {x^3+x-1)^1/2) + x^3+x)^1/2)}

Divide by  So (2x-1)^1/2 is a common factor.

1 = (2x-1)^1/2 / {x^3+x-1)^1/2) +(x^3-1)^1/2) }

Therefore

{x^3+x-1)^(1/2) +(x^3-1)^1/2)} = (2x-1)^1/2) and

{x^3+x-1)^1/2 -(x^3-x)^(1/2) = (2x-1)^(1/2)

Subtracting, 2(x^3 -x)^1/2 =  0 Square both sides

4(x^3-x) =  0

4x(x^2-1) = 0. Divide by 4.

x(x^2-1) = 0

x(x-1)(x+1) = -

x =0, x=1 x=-1 are the solution. In addition we reduced  both sides by 2x-1. So 2x=1 or x = 1/2 is also asolution.

 

 

 

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thewriter | College Teacher | (Level 1) Valedictorian

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The equation (x^3-x)^1/2+(2x-1)^1/2=(x^3+x-1)^1/2 can be solved by taking the square of both the sides. We get:

(x^3-x)+(2x-1)+2*(x^3-x)(2x-1)=(x^3+x-1)

=>x^3-x+2x-1+2(2x^4-2x^2-x^3+x)=x^3+x-1

=>x^3+x-1+4x^4-4x^2-2x^3+2x=x^3+x-1

=>4x^4-4x^2-2x^3+2x=0

=>2x^3-2x-x^2+1=0

=>2x(x^2-1)-1(x^2-1)=0

=>(2x-1)(x^2-1)=0

=>(2x-1)(x-1)(x+1)=0

This gives three values of x which are 1/2, 1, -1

x has the values 1/2, 1, -1

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jt123qwerty | Student, Grade 9 | eNotes Newbie

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http://www.wolframalpha.com/input/?i=%28x^3-x%29^1%2F2+%2B+%282x-1%29^1%2F2+%3D+%28x^3+%2B+x+-+1%29^1%2F2

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